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Design of Structural Elements

Third Edition

Concrete, steelwork, masonry and timber designs toBritish Standards and Eurocodes

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Design ofStructuralElementsThird Edition

Concrete, steelwork, masonry andtimber designs to British Standardsand Eurocodes

Chanakya Arya

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First published 1994 by E & FN Spon

Second edition published 2003 by Spon Press

This edition published 2009by Taylor & Francis2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN

Simultaneously published in the USA and Canadaby Taylor & Francis270 Madison Avenue, New York, NY 10016, USA

Taylor & Francis is an imprint of the Taylor & Francis Group, an informa business

© 1994, 2003, 2009 Chanakya Arya

All rights reserved. No part of this book may be reprinted or reproduced orutilised in any form or by any electronic, mechanical, or other means, nowknown or hereafter invented, including photocopying and recording, or inany information storage or retrieval system, without permission in writingfrom the publishers.

The publisher makes no representation, express or implied, with regardto the accuracy of the information contained in this book and cannot accept anylegal responsibility or liability for any errors or omissions that may be made.

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging in Publication DataArya, Chanakya.Design of structural elements : concrete, steelwork, masonry, and timber designsto British standards and Eurocodes / Chanakya Arya. – 3rd ed.

p. cm.Includes bibliographical references and index.1. Structural design – Standards – Great Britain. 2. Structural design – Standards –Europe. I. Title. II. Title: Concrete, steelwork, masonry, and timber designto British standards and Eurocodes.TA658.A79 2009624.1′7–dc222008043080

ISBN10: 0-415-46719-5 (hbk)ISBN10: 0-415-46720-9 (pbk)ISBN10: 0-203-92650-1 (ebk)

ISBN13: 978-0-415-46719-3 (hbk)ISBN13: 978-0-415-46720-9 (pbk)ISBN13: 978-0-203-92650-5 (ebk)

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This edition published in the Taylor & Francis e-Library, 2009.

To purchase your own copy of this or any of Taylor & Francis or Routledge’scollection of thousands of eBooks please go to www.eBookstore.tandf.co.uk.

ISBN 0-203-92650-1 Master e-book ISBN

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Contents

Preface to the third edition viiPreface to the second edition ixPreface to the first edition xiAcknowledgements xiiiList of worked examples xv

PART ONE: INTRODUCTION TOSTRUCTURAL DESIGN

1 Philosophy of design 31.1 Introduction 31.2 Basis of design 41.3 Summary 8

Questions 8

2 Basic structural concepts andmaterial properties 9

2.1 Introduction 92.2 Design loads acting on structures 92.3 Design loads acting on elements 132.4 Structural analysis 172.5 Beam design 242.6 Column design 262.7 Summary 27

Questions 28

PART TWO: STRUCTURAL DESIGN TOBRITISH STANDARDS

3 Design in reinforced concreteto BS 8110 31

3.1 Introduction 313.2 Objectives and scope 313.3 Symbols 323.4 Basis of design 333.5 Material properties 333.6 Loading 353.7 Stress–strain curves 363.8 Durability and fire resistance 373.9 Beams 44

3.10 Slabs 933.11 Foundations 1153.12 Retaining walls 121

3.13 Design of short braced columns 1283.14 Summary 143

Questions 143

4 Design in structural steelworkto BS 5950 145

4.1 Introduction 1454.2 Iron and steel 1454.3 Structural steel and steel

sections 1464.4 Symbols 1484.5 General principles and design

methods 1494.6 Loading 1504.7 Design strengths 1514.8 Design of steel beams and joists 1514.9 Design of compression members 177

4.10 Floor systems for steel framedstructures 199

4.11 Design of connections 2184.12 Summary 236

Questions 237

5 Design in unreinforced masonryto BS 5628 239

5.1 Introduction 2395.2 Materials 2405.3 Masonry design 2455.4 Symbols 2455.5 Design of vertically loaded masonry

walls 2465.6 Design of laterally loaded wall

panels 2635.7 Summary 276

Questions 277

6 Design in timber to BS 5268 2796.1 Introduction 2796.2 Stress grading 2806.3 Grade stress and strength class 2806.4 Permissible stresses 282

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Contents

6.5 Timber design 2856.6 Symbols 2856.7 Flexural members 2876.8 Design of compression members 2986.9 Design of stud walls 303

6.10 Summary 305Questions 306

PART THREE: STRUCTURAL DESIGNTO THE EUROCODES

7 The structural Eurocodes:An introduction 309

7.1 Scope 3097.2 Benefits of Eurocodes 3097.3 Production of Eurocodes 3107.4 Format 3107.5 Problems associated with drafting

the Eurocodes 3107.6 Decimal point 3127.7 Implementation 3127.8 Maintenance 3127.9 Difference between national

standards and Eurocodes 312

8 Eurocode 2: Design of concretestructures 314

8.1 Introduction 3148.2 Structure of EC 2 3158.3 Symbols 3158.4 Material properties 3168.5 Actions 3178.6 Stress–strain diagrams 3238.7 Cover, fire, durability and bond 3248.8 Design of singly and doubly

reinforced rectangular beams 3278.9 Design of one-way solid slabs 350

8.10 Design of pad foundations 3578.11 Design of columns 361

9 Eurocode 3: Design of steelstructures 375

9.1 Introduction 3759.2 Structure of EC 3 3769.3 Principles and Application rules 3769.4 Nationally Determined

Parameters 3769.5 Symbols 3779.6 Member axes 3779.7 Basis of design 377

9.8 Actions 3789.9 Materials 378

9.10 Classification of cross-sections 3809.11 Design of beams 3809.12 Design of columns 4039.13 Connections 418

10 Eurocode 6: Design ofmasonry structures 434

10.1 Introduction 43410.2 Layout 43410.3 Principles/Application rules 43510.4 Nationally Determined

Parameters 43510.5 Symbols 43510.6 Basis of design 43610.7 Actions 43610.8 Design compressive strength 43710.9 Durability 441

10.10 Design of unreinforced masonrywalls subjected to verticalloading 441

10.11 Design of laterally loaded wallpanels 455

11 Eurocode 5: Design oftimber structures 458

11.1 Introduction 45811.2 Layout 45811.3 Principles/Application rules 45911.4 Nationally Determined

Parameters 45911.5 Symbols 45911.6 Basis of design 46011.7 Design of flexural members 46411.8 Design of columns 477

Appendix A Permissible stress and loadfactor design 481

Appendix B Dimensions and propertiesof steel universal beams andcolumns 485

Appendix C Buckling resistance ofunstiffened webs 489

Appendix D Second moment of area of acomposite beam 491

Appendix E References and furtherreading 493

Index 497

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Preface to thethird edition

Since publication of the second edition of Designof Structural Elements there have been two majordevelopments in the field of structural engineeringwhich have suggested this new edition.

The first and foremost of these is that theEurocodes for concrete, steel, masonry and timberdesign have now been converted to full EuroNorm(EN) status and, with the possible exception of thesteel code, all the associated UK National Annexeshave also been finalised and published. Therefore,these codes can now be used for structural design,although guidance on the timing and circumstancesunder which they must be used is still awaited.Thus, the content of Chapters 8 to 11 on, respec-tively, the design of concrete, steel, masonry andtimber structures has been completely revised tocomply with the EN versions of the Eurocodes forthese materials. The opportunity has been used toexpand Chapter 10 and include several workedexamples on the design of masonry walls subject toeither vertical or lateral loading or a combinationof both.

The second major development is that a numberof small but significant amendments have been

made to the 1997 edition of BS 8110: Part 1 onconcrete design, and new editions of BS 5628:Parts 1 and 3 on masonry design have recentlybeen published. These and other national stand-ards, e.g. BS 5950 for steel design and BS 5268for timber design, are still widely used in the UKand beyond. This situation is likely to persist forsome years, and therefore the decision was takento retain the chapters on British Standards andwhere necessary update the material to reflect latestdesign recommendations. This principally affectsthe material in Chapters 3 and 5 on concrete andmasonry design.

The chapters on Eurocodes are not self-containedbut include reference to relevant chapters on BritishStandards. This should not present any problemsto readers familiar with British Standards, but willmean that readers new to this subject will have torefer to two chapters from time to time to get themost from this book. This is not ideal, but shouldresult in the reader becoming familiar with bothBritish and European practices, which is probablynecessary during the transition phase from BritishStandards to Eurocodes.

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Preface to thesecond edition

The main motivation for preparing this newedition was to update the text in Chapters 4 and 6on steel and timber design to conform with thelatest editions of respectively BS 5950: Part 1 andBS 5268: Part 2. The opportunity has also beentaken to add new material to Chapters 3 and 4.Thus, Chapter 3 on concrete design now includesa new section and several new worked examples onthe analysis and design of continuous beams andslabs. Examples illustrating the analysis and designof two-way spanning slabs and columns subjectto axial load and bending have also been added.The section on concrete slabs has been updated. Adiscussion on flooring systems for steel framedstructures is featured in Chapter 4 together with asection and several worked examples on compositefloor design.

Work on converting Parts 1.1 of the Eurocodesfor concrete, steel, timber and masonry structures

to full EN status is still ongoing. Until such timethat these documents are approved the design rulesin pre-standard form, designated by ENV, remainvalid. The material in Chapters 8, 9 and 11 tothe ENV versions of EC2, EC3 and EC5 are stillcurrent. The first part of Eurocode 6 on masonrydesign was published in pre-standard form in1996, some three years after publication of the firstedition of this book. The material in Chapter 10has therefore been revised, so it now conforms tothe guidance given in the ENV.

I would like to thank the following who haveassisted with the preparation of this new edition: Pro-fessor Colin Baley for preparing Appendix C; FredLambert, Tony Threlfall, Charles Goodchild andPeter Watt for reviewing parts of the manuscript.

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Preface to thefirst edition

individual elements can be assessed,thereby enabling the designer to sizethe element.

Part Two contains four chapters covering thedesign and detailing of a number ofstructural elements, e.g. floors, beams,walls, columns, connections andfoundations to the latest British codesof practice for concrete, steelwork,masonry and timber design.

Part Three contains five chapters on the Euro-codes for these materials. The first ofthese describes the purpose, scope andproblems associated with drafting theEurocodes. The remaining chaptersdescribe the layout and contents ofEC2, EC3, EC5 and EC6 for designin concrete, steelwork, timber andmasonry respectively.

At the end of Chapters 1–6 a number of designproblems have been included for the student toattempt.

Although most of the tables and figures fromthe British Standards referred to in the text havebeen reproduced, it is expected that the readerwill have either the full Standard or the publica-tion Extracts from British Standards for Students ofStructural Design in order to gain the most fromthis book.

I would like to thank the following who haveassisted with the production of this book: PeterWright for co-authoring Chapters 1, 4 and 9; FredLambert, Tony Fewell, John Moran, David Smith,Tony Threlfall, Colin Taylor, Peter Watt and PeterSteer for reviewing various parts of the manuscript;Tony Fawcett for the drafting of the figures;and Associate Professor Noor Hana for help withproofreading.

C. AryaLondon

UK

Structural design is a key element of all degree anddiploma courses in civil and structural engineering.It involves the study of principles and procedurescontained in the latest codes of practice for struc-tural design for a range of materials, including con-crete, steel, masonry and timber.

Most textbooks on structural design consider onlyone construction material and, therefore, the studentmay end up buying several books on the subject.This is undesirable from the viewpoint of cost butalso because it makes it difficult for the studentto unify principles of structural design, because ofdiffering presentation approaches adopted by theauthors.

There are a number of combined textbooks whichinclude sections on several materials. However,these tend to concentrate on application of thecodes and give little explanation of the structuralprinciples involved or, indeed, an awareness ofmaterial properties and their design implications.Moreover, none of the books refer to the newEurocodes for structural design, which will eventu-ally replace British Standards.

The purpose of this book, then, is to describethe background to the principles and procedurescontained in the latest British Standards andEurocodes on the structural use of concrete, steel-work, masonry and timber. It is primarily aimed atstudents on civil and structural engineering degreeand diploma courses. Allied professionals such asarchitects, builders and surveyors will also find itappropriate. In so far as it includes five chapters onthe structural Eurocodes it will be of considerableinterest to practising engineers too.

The subject matter is divided into 11 chaptersand 3 parts:

Part One contains two chapters and explains theprinciples and philosophy of structuraldesign, focusing on the limit stateapproach. It also explains how theoverall loading on a structure and

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Acknowledgements

I am once again indebted to Tony Threlfall, for-merly of the British Cement Association and nowan independent consultant, for comprehensively re-viewing Chapter 8 and the material in Chapter 3on durability and fire resistance

I would also sincerely like to thank ProfessorR.S. Narayanan of the Clark Smith Partnershipfor reviewing Chapter 7, David Brown of theSteel Construction Institute for reviewing Chap-ter 9, Dr John Morton, an independent consultant,for reviewing Chapter 10, Dr Ali Arasteh of theBrick Development Association for reviewing Chap-ters 5 and 10, and Peter Steer, an independent

consultant, for reviewing Chapter 11. The contentsof these chapters are greatly improved due to theircomments.

A special thanks to John Aston for reading partsof the manuscript.

I am grateful to The Concrete Centre for per-mission to use extracts from their publications.Extracts from British Standards are reproduced withthe permission of BSI under licence number2008ET0037. Complete standards can be obtainedfrom BSI Customer Services, 389 Chiswick HighRoad, London W4 4AL.

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List of workedexamples

2.1 Self-weight of a reinforced concretebeam 10

2.2 Design loads on a floor beam 142.3 Design loads on floor beams and

columns 152.4 Design moments and shear forces in

beams using equilibrium equations 182.5 Design moments and shear forces in

beams using formulae 232.6 Elastic and plastic moments of

resistance of a beam section 262.7 Analysis of column section 27

3.1 Selection of minimum strength classand nominal concrete cover toreinforcement (BS 8110) 43

3.2 Design of bending reinforcement fora singly reinforced beam (BS 8110) 48

3.3 Design of shear reinforcement for abeam (BS 8110) 52

3.4 Sizing a concrete beam (BS 8110) 593.5 Design of a simply supported

concrete beam (BS 8110) 613.6 Analysis of a singly reinforced

concrete beam (BS 8110) 653.7 Design of bending reinforcement for

a doubly reinforced beam (BS 8110) 683.8 Analysis of a two-span continuous

beam using moment distribution 723.9 Analysis of a three span continuous

beam using moment distribution 763.10 Continuous beam design (BS 8110) 783.11 Design of a one-way spanning

concrete floor (BS 8110) 1003.12 Analysis of a one-way spanning

concrete floor (BS 8110) 1043.13 Continuous one-way spanning slab

design (BS 8110) 1063.14 Design of a two-way spanning

restrained slab (BS 8110) 1103.15 Design of a pad footing (BS8110) 117

3.16 Design of a cantilever retaining wall(BS 8110) 125

3.17 Classification of a concrete column(BS 8110) 131

3.18 Sizing a concrete column (BS 8110) 1333.19 Analysis of a column section

(BS 8110) 1343.20 Design of an axially loaded column

(BS 8110) 1393.21 Column supporting an approximately

symmetrical arrangement of beams(BS 8110) 140

3.22 Columns resisting an axial load andbending (BS 8110) 141

4.1 Selection of a beam section inS275 steel (BS 5950) 156

4.2 Selection of beam section inS460 steel (BS 5950) 158

4.3 Selection of a cantilever beamsection (BS 5950) 159

4.4 Deflection checks on steel beams(BS 5950) 161

4.5 Checks on web bearing and bucklingfor steel beams (BS 5950) 164

4.6 Design of a steel beam with webstiffeners (BS 5950) 164

4.7 Design of a laterally unrestrained steelbeam – simple method (BS 5950) 171

4.8 Design of a laterally unrestrainedbeam – rigorous method (BS 5950) 174

4.9 Checking for lateral instability in acantilever steel beam (BS 5950) 176

4.10 Design of an axially loaded column(BS 5950) 183

4.11 Column resisting an axial load andbending (BS 5950) 185

4.12 Design of a steel column in ‘simple’construction (BS 5950) 189

4.13 Encased steel column resisting an axialload (BS 5950) 193

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4.14 Encased steel column resisting anaxial load and bending (BS 5950) 195

4.15 Design of a steel column baseplate(BS 5950) 198

4.16 Advantages of compositeconstruction (BS 5950) 200

4.17 Moment capacity of a compositebeam (BS 5950) 209

4.18 Moment capacity of a compositebeam (BS 5950) 210

4.19 Design of a composite floor(BS 5950) 212

4.20 Design of a composite floorincorporating profiled metal decking(BS 5950) 215

4.21 Beam-to-column connection usingweb cleats (BS 5950) 224

4.22 Analysis of a bracket-to-columnconnection (BS 5950) 227

4.23 Analysis of a beam splice connection(BS 5950) 228

4.24 Analysis of a beam-to-columnconnection using an end plate(BS 5950) 232

4.25 Analysis of a welded beam-to-columnconnection (BS 5950) 235

5.1 Design of a load-bearing brick wall(BS 5628) 254

5.2 Design of a brick wall with ‘small’plan area (BS 5628) 255

5.3 Analysis of brick walls stiffened withpiers (BS 5628) 256

5.4 Design of single leaf brick and blockwalls (BS 5628) 258

5.5 Design of a cavity wall (BS 5628) 2615.6 Analysis of a one-way spanning wall

panel (BS 5628) 2715.7 Analysis of a two-way spanning panel

wall (BS 5628) 2725.8 Design of a two-way spanning

single-leaf panel wall (BS 5628) 2735.9 Analysis of a two-way spanning

cavity panel wall (BS 5628) 274

6.1 Design of a timber beam (BS 5268) 2916.2 Design of timber floor joists

(BS 5268) 2936.3 Design of a notched floor joist

(BS 5268) 2966.4 Analysis of a timber roof (BS 5268) 2966.5 Timber column resisting an axial

load (BS 5268) 301

6.6 Timber column resisting an axialload and moment (BS 5268) 302

6.7 Analysis of a stud wall (BS 5268) 304

8.1 Design actions for simply supportedbeam (EN 1990) 321

8.2 Bending reinforcement for a singlyreinforced beam (EC 2) 329

8.3 Bending reinforcement for a doublyreinforced beam (EC 2) 329

8.4 Design of shear reinforcement for abeam (EC 2) 334

8.5 Design of shear reinforcement atbeam support (EC 2) 335

8.6 Deflection check for concrete beams(EC 2) 338

8.7 Calculation of anchorage lengths(EC 2) 342

8.8 Design of a simply supportedbeam (EC 2) 345

8.9 Analysis of a singly reinforcedbeam (EC 2) 349

8.10 Design of a one-way spanningfloor (EC 2) 352

8.11 Analysis of one-way spanningfloor (EC 2) 355

8.12 Design of a pad foundation (EC 2) 3598.13 Column supporting an axial load

and uni-axial bending (EC 2) 3668.14 Classification of a column (EC 2) 3678.15 Classification of a column (EC 2) 3698.16 Column design (i) λ < λlim;

(ii) λ > λlim (EC 2) 3708.17 Column subjected to combined

axial load and biaxial bending(EC 2) 373

9.1 Analysis of a laterally restrainedbeam (EC 3) 384

9.2 Design of a laterally restrainedbeam (EC 3) 387

9.3 Design of a cantilever beam(EC 3) 391

9.4 Design of a beam with stiffeners(EC 3) 393

9.5 Analysis of a beam restrained at thesupports (EC 3) 401

9.6 Analysis of a beam restrained atmid-span and supports (EC 3) 402

9.7 Analysis of a column resisting anaxial load (EC 3) 408

9.8 Analysis of a column with a tie-beamat mid-height (EC 3) 410

List of worked examples

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9.9 Analysis of a column resisting anaxial load and moment (EC 3) 411

9.10 Analysis of a steel column in ‘simple’construction (EC 3) 415

9.11 Analysis of a column baseplate(EC 3) 417

9.12 Analysis of a tension splice connection(EC 3) 422

9.13 Shear resistance of a welded end plateto beam connection (EC 3) 424

9.14 Bolted beam-to-column connectionusing an end plate (EC 3) 426

9.15 Bolted beam-to-column connectionusing web cleats (EC 3) 429

10.1 Design of a loadbearing brickwall (EC 6) 444

10.2 Design of a brick wall with ‘small’plan area (EC 6) 446

10.3 Analysis of brick walls stiffened withpiers (EC 6) 447

10.4 Design of a cavity wall (EC 6) 45010.5 Block wall subject to axial load

and wind (EC 6) 45310.6 Analysis of a one-way spanning

wall panel (EC 6) 45610.7 Analysis of a two-way spanning

panel wall (EC 6) 457

11.1 Design of timber floor joists (EC 5) 46911.2 Design of a notched floor joist

(EC 5) 47511.3 Analysis of a solid timber beam

restrained at supports (EC 5) 47611.4 Analysis of a column resisting an

axial load (EC 5) 47811.5 Analysis of an eccentrically loaded

column (EC 5) 479

List of worked examples

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DedicationIn memory of Biji

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PART ONE

INTRODUCTIONTO STRUCTURALDESIGN

The primary aim of all structural design is toensure that the structure will perform satisfactorilyduring its design life. Specifically, the designer mustcheck that the structure is capable of carrying theloads safely and that it will not deform excessivelydue to the applied loads. This requires the de-signer to make realistic estimates of the strengths ofthe materials composing the structure and the load-ing to which it may be subject during its designlife. Furthermore, the designer will need a basicunderstanding of structural behaviour.

The work that follows has two objectives:

1. to describe the philosophy of structural design;2. to introduce various aspects of structural and

material behaviour.

Towards the first objective, Chapter 1 discusses thethree main philosophies of structural design, emphas-izing the limit state philosophy which forms the basesof design in many of the modern codes of practice.Chapter 2 then outlines a method of assessing thedesign loading acting on individual elements of astructure and how this information can be used, to-gether with the material properties, to size elements.

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Philosophy of design

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Chapter 1


This chapter is concerned with the philosophy of struc-tural design. The chapter describes the overall aims ofdesign and the many inputs into the design process.The primary aim of design is seen as the need to ensurethat at no point in the structure do the design loadsexceed the design strengths of the materials. This can beachieved by using the permissible stress or load factorphilosophies of design. However, both suffer from draw-backs and it is more common to design according tolimit state principles which involve considering all themechanisms by which a structure could become unfitfor its intended purpose during its design life.

1.1 IntroductionThe task of the structural engineer is to design astructure which satisfies the needs of the client andthe user. Specifically the structure should be safe,economical to build and maintain, and aesthetic-ally pleasing. But what does the design processinvolve?

Design is a word that means different things todifferent people. In dictionaries the word is de-scribed as a mental plan, preliminary sketch, pat-tern, construction, plot or invention. Even amongthose closely involved with the built environmentthere are considerable differences in interpretation.Architects, for example, may interpret design asbeing the production of drawings and models toshow what a new building will actually look like.To civil and structural engineers, however, design istaken to mean the entire planning process for a newbuilding structure, bridge, tunnel, road, etc., fromoutline concepts and feasibility studies throughmathematical calculations to working drawingswhich could show every last nut and bolt in theproject. Together with the drawings there will bebills of quantities, a specification and a contract,which will form the necessary legal and organiza-tional framework within which a contractor, under

the supervision of engineers and architects, can con-struct the scheme.

There are many inputs into the engineeringdesign process as illustrated by Fig. 1.1 including:

1. client brief2. experience3. imagination4. a site investigation5. model and laboratory tests6. economic factors7. environmental factors.

The starting-point for the designer is normallya conceptual brief from the client, who may be aprivate developer or perhaps a government body.The conceptual brief may simply consist of somesketches prepared by the client or perhaps a detailedset of architect’s drawings. Experience is cruciallyimportant, and a client will always demand thatthe firm he is employing to do the design has pre-vious experience designing similar structures.

Although imagination is thought by some tobe entirely the domain of the architect, this is notso. For engineers and technicians an imaginationof how elements of structure interrelate in threedimensions is essential, as is an appreciation ofthe loadings to which structures might be subjectin certain circumstances. In addition, imaginativesolutions to engineering problems are often requiredto save money, time, or to improve safety or quality.

A site investigation is essential to determine thestrength and other characteristics of the groundon which the structure will be founded. If the struc-ture is unusual in any way, or subject to abnormalloadings, model or laboratory tests may also be usedto help determine how the structure will behave.

In today’s economic climate a structural designermust be constantly aware of the cost implicationsof his or her design. On the one hand design shouldaim to achieve economy of materials in the struc-ture, but over-refinement can lead to an excessive

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number of different sizes and components in thestructure, and labour costs will rise. In additionthe actual cost of the designer’s time should not beexcessive, or this will undermine the employer’scompetitiveness. The idea is to produce a workabledesign achieving reasonable economy of materials,while keeping manufacturing and construction costsdown, and avoiding unnecessary design and researchexpenditure. Attention to detailing and buildabilityof structures cannot be overemphasized in design.Most failures are as a result of poor detailing ratherthan incorrect analysis.

Designers must also understand how the struc-ture will fit into the environment for which it isdesigned. Today many proposals for engineeringstructures stand or fall on this basis, so it is part ofthe designer’s job to try to anticipate and recon-cile the environmental priorities of the public andgovernment.

The engineering design process can often bedivided into two stages: (1) a feasibility study in-volving a comparison of the alternative forms ofstructure and selection of the most suitable type and(2) a detailed design of the chosen structure. Thesuccess of stage 1, the conceptual design, reliesto a large extent on engineering judgement andinstinct, both of which are the outcome of manyyears’ experience of designing structures. Stage 2,the detailed design, also requires these attributesbut is usually more dependent upon a thoroughunderstanding of the codes of practice for struc-tural design, e.g. BS 8110 and BS 5950. Thesedocuments are based on the amassed experience of

many generations of engineers, and the results ofresearch. They help to ensure safety and economyof construction, and that mistakes are not repeated.For instance, after the infamous disaster at theRonan Point block of flats in Newham, London,when a gas explosion caused a serious partial col-lapse, research work was carried out, and codes ofpractice were amended so that such structures couldsurvive a gas explosion, with damage being con-fined to one level.

The aim of this book is to look at the proceduresassociated with the detailed design of structuralelements such as beams, columns and slabs. Chap-ter 2 will help the reader to revise some basic the-ories of structural behaviour. Chapters 3–6 deal withdesign to British Standard (BS) codes of practicefor the structural use of concrete (BS 8110), struc-tural steelwork (BS 5950), masonry (BS 5628) andtimber (BS 5268). Chapter 7 introduces the newEurocodes (EC) for structural design and Chapters8–11 then describe the layout and design principlesin EC2, EC3, EC6 and EC5 for concrete, steel-work, masonry and timber respectively.

1.2 Basis of designTable 1.1 illustrates some risk factors that are asso-ciated with activities in which people engage. Itcan be seen that some degree of risk is associatedwith air and road travel. However, people normallyaccept that the benefits of mobility outweigh therisks. Staying in buildings, however, has always been

Fig. 1.1 Inputs into the design process.

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critical points, as stress due to loading exceeds thestrength of the material. In order for the structureto be safe the overlapping area must be kept to aminimum. The degree of overlap between the twocurves can be minimized by using one of three dis-tinct design philosophies, namely:

1. permissible stress design2. load factor method3. limit state design.

1.2.1 PERMISSIBLE STRESS DESIGNIn permissible stress design, sometimes referred toas modular ratio or elastic design, the stresses in thestructure at working loads are not allowed to exceeda certain proportion of the yield stress of the con-struction material, i.e. the stress levels are limitedto the elastic range. By assuming that the stress–strain relationship over this range is linear, it is pos-sible to calculate the actual stresses in the materialconcerned. Such an approach formed the basis of thedesign methods used in CP 114 (the forerunner ofBS 8110) and BS 449 (the forerunner of BS 5950).

However, although it modelled real building per-formance under actual conditions, this philosophyhad two major drawbacks. Firstly, permissible designmethods sometimes tended to overcomplicate thedesign process and also led to conservative solutions.Secondly, as the quality of materials increased andthe safety margins decreased, the assumption thatstress and strain are directly proportional becameunjustifiable for materials such as concrete, makingit impossible to estimate the true factors of safety.

1.2.2 LOAD FACTOR DESIGNLoad factor or plastic design was developed to takeaccount of the behaviour of the structure once theyield point of the construction material had beenreached. This approach involved calculating thecollapse load of the structure. The working load wasderived by dividing the collapse load by a load factor.This approach simplified methods of analysis andallowed actual factors of safety to be calculated.It was in fact permitted in CP 114 and BS 449but was slow in gaining acceptance and was even-tually superseded by the more comprehensive limitstate approach.

The reader is referred to Appendix A for an ex-ample illustrating the differences between the per-missible stress and load factor approaches to design.

1.2.3 LIMIT STATE DESIGNOriginally formulated in the former Soviet Unionin the 1930s and developed in Europe in the 1960s,

Table 1.1 Comparative death risk per 108

persons exposed

Mountaineering (international) 2700Air travel (international) 120Deep water trawling 59Car travel 56Coal mining 21Construction sites 8Manufacturing 2Accidents at home 2Fire at home 0.1Structural failures 0.002

Fig. 1.2 Relationship between stress and strength.

regarded as fairly safe. The risk of death or injurydue to structural failure is extremely low, but as wespend most of our life in buildings this is perhapsjust as well.

As far as the design of structures for safety isconcerned, it is seen as the process of ensuringthat stresses due to loading at all critical points in astructure have a very low chance of exceeding thestrength of materials used at these critical points.Figure 1.2 illustrates this in statistical terms.

In design there exist within the structure a numberof critical points (e.g. beam mid-spans) where thedesign process is concentrated. The normal distribu-tion curve on the left of Fig. 1.2 represents the actualmaximum material stresses at these critical pointsdue to the loading. Because loading varies accordingto occupancy and environmental conditions, andbecause design is an imperfect process, the materialstresses will vary about a modal value – the peak ofthe curve. Similarly the normal distribution curveon the right represents material strengths at thesecritical points, which are also not constant due tothe variability of manufacturing conditions.

The overlap between the two curves represents apossibility that failure may take place at one of the

Basis of design

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limit state design can perhaps be seen as a com-promise between the permissible and load factormethods. It is in fact a more comprehensive ap-proach which takes into account both methods inappropriate ways. Most modern structural codes ofpractice are now based on the limit state approach.BS 8110 for concrete, BS 5950 for structuralsteelwork, BS 5400 for bridges and BS 5628 formasonry are all limit state codes. The principalexceptions are the code of practice for design intimber, BS 5268, and the old (but still current)structural steelwork code, BS 449, both of whichare permissible stress codes. It should be noted, how-ever, that the Eurocode for timber (EC5), which isexpected to replace BS 5268 around 2010, is basedon limit state principles.

As limit state philosophy forms the basis of thedesign methods in most modern codes of practicefor structural design, it is essential that the designmethodology is fully understood. This then is thepurpose of the following subsections.

1.2.3.1 Ultimate and serviceabilitylimit states

The aim of limit state design is to achieve accept-able probabilities that a structure will not becomeunfit for its intended use during its design life, thatis, the structure will not reach a limit state. Thereare many ways in which a structure could becomeunfit for use, including excessive conditions of bend-ing, shear, compression, deflection and cracking(Fig. 1.3). Each of these mechanisms is a limit statewhose effect on the structure must be individuallyassessed.

Some of the above limit states, e.g. deflectionand cracking, principally affect the appearance ofthe structure. Others, e.g. bending, shear and com-pression, may lead to partial or complete collapseof the structure. Those limit states which can causefailure of the structure are termed ultimate limitstates. The others are categorized as serviceabilitylimit states. The ultimate limit states enable thedesigner to calculate the strength of the structure.Serviceability limit states model the behaviour of thestructure at working loads. In addition, there maybe other limit states which may adversely affectthe performance of the structure, e.g. durabilityand fire resistance, and which must therefore alsobe considered in design.

It is a matter of experience to be able to judgewhich limit states should be considered in thedesign of particular structures. Nevertheless, oncethis has been done, it is normal practice to base

the design on the most critical limit state and thencheck for the remaining limit states. For example,for reinforced concrete beams the ultimate limitstates of bending and shear are used to size thebeam. The design is then checked for the remain-ing limit states, e.g. deflection and cracking. Onthe other hand, the serviceability limit state ofdeflection is normally critical in the design of con-crete slabs. Again, once the designer has determineda suitable depth of slab, he/she must then makesure that the design satisfies the limit states of bend-ing, shear and cracking.

In assessing the effect of a particular limit stateon the structure, the designer will need to assumecertain values for the loading on the structure andthe strength of the materials composing the struc-ture. This requires an understanding of the con-cepts of characteristic and design values which arediscussed below.

1.2.3.2 Characteristic and design valuesAs stated at the outset, when checking whether aparticular member is safe, the designer cannot becertain about either the strength of the materialcomposing the member or, indeed, the load whichthe member must carry. The material strength maybe less than intended (a) because of its variablecomposition, and (b) because of the variability ofmanufacturing conditions during construction, andother effects such as corrosion. Similarly the loadin the member may be greater than anticipated (a)because of the variability of the occupancy or envir-onmental loading, and (b) because of unforeseencircumstances which may lead to an increase in thegeneral level of loading, errors in the analysis, errorsduring construction, etc.

In each case, item (a) is allowed for by using acharacteristic value. The characteristic strengthis the value below which the strength lies in only asmall number of cases. Similarly the characteristicload is the value above which the load lies in onlya small percentage of cases. In the case of strengththe characteristic value is determined from test re-sults using statistical principles, and is normallydefined as the value below which not more than5% of the test results fall. However, at this stagethere are insufficient data available to apply statist-ical principles to loads. Therefore the characteristicloads are normally taken to be the design loadsfrom other codes of practice, e.g. BS 648 and BS6399.

The overall effect of items under (b) is allowedfor using a partial safety factor: γm for strength

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Fig. 1.3 Typical modes of failure for beams and columns.

Basis of design

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and γ f for load. The design strength is obtained bydividing the characteristic strength by the partialsafety factor for strength:

Design strength

characteristic strength

m

=γ

(1.1)

The design load is obtained by multiplying thecharacteristic load by the partial safety factor forload:

Design load = characteristic load × γ f (1.2)

The value of γm will depend upon the propertiesof the actual construction material being used.Values for γ f depend on other factors which will bediscussed more fully in Chapter 2.

In general, once a preliminary assessment of thedesign loads has been made it is then possible tocalculate the maximum bending moments, shearforces and deflections in the structure (Chapter 2).The construction material must be capable ofwithstanding these forces otherwise failure of thestructure may occur, i.e.

Design strength ≥ design load (1.3)

Simplified procedures for calculating the moment,shear and axial load capacities of structural ele-ments together with acceptable deflection limitsare described in the appropriate codes of practice.

These allow the designer to rapidly assess the suit-ability of the proposed design. However, beforediscussing these procedures in detail, Chapter 2describes in general terms how the design loadsacting on the structure are estimated and used tosize individual elements of the structure.

1.3 SummaryThis chapter has examined the bases of threephilosophies of structural design: permissible stress,load factor and limit state. The chapter has con-centrated on limit state design since it forms thebasis of the design methods given in the codes ofpractice for concrete (BS 8110), structural steel-work (BS 5950) and masonry (BS 5628). The aimof limit state design is to ensure that a structurewill not become unfit for its intended use, that is,it will not reach a limit state during its design life.Two categories of limit states are examined indesign: ultimate and serviceability. The former isconcerned with overall stability and determiningthe collapse load of the structure; the latter exam-ines its behaviour under working loads. Structuraldesign principally involves ensuring that the loadsacting on the structure do not exceed its strengthand the first step in the design process then is toestimate the loads acting on the structure.

Questions1. Explain the difference between conceptual

design and detailed design.2. What is a code of practice and what is its

purpose in structural design?3. List the principal sources of uncertainty in

structural design and discuss how theseuncertainties are rationally allowed for indesign.

4. The characteristic strengths and designstrengths are related via the partial safetyfactor for materials. The partial safetyfactor for concrete is higher than for steelreinforcement. Discuss why this should be so.

5. Describe in general terms the ways inwhich a beam and column could becomeunfit for use.

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Chapter 2

Basic structuralconcepts andmaterial properties

This chapter is concerned with general methods ofsizing beams and columns in structures. The chapterdescribes how the characteristic and design loads actingon structures and on the individual elements are deter-mined. Methods of calculating the bending moments,shear forces and deflections in beams are outlined.Finally, the chapter describes general approaches tosizing beams according to elastic and plastic criteriaand sizing columns subject to axial loading.

2.1 IntroductionAll structures are composed of a number of inter-connected elements such as slabs, beams, columns,walls and foundations. Collectively, they enable theinternal and external loads acting on the structureto be safely transmitted down to the ground. Theactual way that this is achieved is difficult to modeland many simplifying, but conservative, assump-tions have to be made. For example, the degreeof fixity at column and beam ends is usually uncer-tain but, nevertheless, must be estimated as itsignificantly affects the internal forces in the element.Furthermore, it is usually assumed that the reactionfrom one element is a load on the next and thatthe sequence of load transfer between elementsoccurs in the order: ceiling/floor loads to beams tocolumns to foundations to ground (Fig. 2.1).

At the outset, the designer must make an assess-ment of the future likely level of loading, includingself-weight, to which the structure may be subjectduring its design life. Using computer methods orhand calculations the design loads acting on indi-vidual elements can then be evaluated. The designloads are used to calculate the bending moments,shear forces and deflections at critical points alongthe elements. Finally, suitable dimensions for theelement can be determined. This aspect requiresan understanding of the elementary theory ofbending and the behaviour of elements subject to

Fig. 2.1 Sequence of load transfer between elements of astructure.

compressive loading. These steps are summarizedin Fig. 2.2 and the following sections describe theprocedures associated with each step.

2.2 Design loads acting onstructures

The loads acting on a structure are divided intothree basic types: dead, imposed and wind. Foreach type of loading there will be characteristic anddesign values, as discussed in Chapter 1, which mustbe estimated. In addition, the designer will have todetermine the particular combination of loadingwhich is likely to produce the most adverse effecton the structure in terms of bending moments,shear forces and deflections.

2.2.1 DEAD LOADS, G k, gkDead loads are all the permanent loads acting onthe structure including self-weight, finishes, fixturesand partitions. The characteristic dead loads can be

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Basic structural concepts and material properties

10

Fig. 2.2 Design process.

Example 2.1 Self-weight of a reinforced concrete beamCalculate the self-weight of a reinforced concrete beam of breadth 300 mm, depth 600 mm and length 6000 mm.

From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming that the gravitational constant is10 m s−2 (strictly 9.807 m s−2), the unit weight of reinforced concrete, ρ, is

ρ = 2400 × 10 = 24 000 N m−3 = 24 kN m−3

Hence, the self-weight of beam, SW, is

SW = volume × unit weight

= (0.3 × 0.6 × 6)24 = 25.92 kN

estimated using the schedule of weights of buildingmaterials given in BS 648 (Table 2.1) or from manu-facturers’ literature. The symbols Gk and gk arenormally used to denote the total and uniformlydistributed characteristic dead loads respectively.

Estimation of the self-weight of an element tendsto be a cyclic process since its value can only beassessed once the element has been designed whichrequires prior knowledge of the self-weight of theelement. Generally, the self-weight of the elementis likely to be small in comparison with other deadand live loads and any error in estimation will tendto have a minimal effect on the overall design(Example 2.1).

2.2.2 IMPOSED LOADS Q k, qkImposed load, sometimes also referred to as liveload, represents the load due to the proposed oc-cupancy and includes the weights of the occupants,furniture and roof loads including snow. Sinceimposed loads tend to be much more variablethan dead loads they are more difficult to predict.

BS 6399: Part 1: 1984: Code of Practice for Dead andImposed Loads gives typical characteristic imposedfloor loads for different classes of structure, e.g.residential dwellings, educational institutions,hospitals, and parts of the same structure, e.g.balconies, corridors and toilet rooms (Table 2.2).

2.2.3 WIND LOADSWind pressure can either add to the other gravita-tional forces acting on the structure or, equallywell, exert suction or negative pressures on thestructure. Under particular situations, the latter maywell lead to critical conditions and must be con-sidered in design. The characteristic wind loadsacting on a structure can be assessed in accordancewith the recommendations given in CP 3: ChapterV: Part 2: 1972 Wind Loads or Part 2 of BS 6399:Code of Practice for Wind Loads.

Wind loading is important in the design of ma-sonry panel walls (Chapter 5). However beyond that,wind loading is not considered further since the em-phasis in this book is on the design of elements rather

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Table 2.1 Schedule of unit masses of building materials (based on BS 648)

AsphaltRoofing 2 layers, 19 mm thick 42 kg m−2

Damp-proofing, 19 mm thick 41 kg m−2

Roads and footpaths, 19 mm thick 44 kg m−2

Bitumen roofing feltsMineral surfaced bitumen 3.5 kg m−2

BlockworkSolid per 25 mm thick, stone 55 kg m−2

aggregateAerated per 25 mm thick 15 kg m−2

BoardBlockboard per 25 mm thick 12.5 kg m−2

BrickworkClay, solid per 25 mm thick 55 kg m−2

medium densityConcrete, solid per 25 mm thick 59 kg m−2

Cast stone 2250 kg m−3

ConcreteNatural aggregates 2400 kg m−3

Lightweight aggregates (structural) 1760 + 240/−160 kg m−3

FlagstonesConcrete, 50 mm thick 120 kg m−2

Glass fibreSlab, per 25 mm thick 2.0–5.0 kg m−2

Gypsum panels and partitionsBuilding panels 75 mm thick 44 kg m−2

LeadSheet, 2.5 mm thick 30 kg m−2

Linoleum3 mm thick 6 kg m−2

loads, γ f (Chapter 1). The value for γ f depends onseveral factors including the limit state underconsideration, i.e. ultimate or serviceability, theaccuracy of predicting the load and the particu-lar combination of loading which will produce theworst possible effect on the structure in terms ofbending moments, shear forces and deflections.

than structures, which generally involves investigat-ing the effects of dead and imposed loads only.

2.2.4 LOAD COMBINATIONS ANDDESIGN LOADS

The design loads are obtained by multiplying thecharacteristic loads by the partial safety factor for

PlasterTwo coats gypsum, 13 mm thick 22 kg m−2

Plastics sheeting (corrugated) 4.5 kg m−2

Plywoodper mm thick 0.7 kg m−2

Reinforced concrete 2400 kg m−3

RenderingCement: sand (1:3), 13 mm thick 30 kg m−2

ScreedingCement: sand (1:3), 13 mm thick 30 kg m−2

Slate tiles(depending upon thickness 24 –78 kg m−3

and source)

SteelSolid (mild) 7850 kg m−3

Corrugated roofing sheets, 10 kg m−2

per mm thick

Tarmacadam25 mm thick 60 kg m−2

Terrazzo25 mm thick 54 kg m−2

Tiling, roofClay 70 kg m−2

TimberSoftwood 590 kg m−3

Hardwood 1250 kg m−3

Water 1000 kg m−3

WoodwoolSlabs, 25 mm thick 15 kg m−2

Design loads acting on structures

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Table 2.2 Imposed loads for residential occupancy class

Floor area usage Intensity of distributed load Concentrated loadkN m−2 kN

Type 1. Self-contained dwelling unitsAll 1.5 1.4

Type 2. Apartment houses, boarding houses, lodginghouses, guest houses, hostels, residential clubs andcommunal areas in blocks of flatsBoiler rooms, motor rooms, fan rooms and the like 7.5 4.5

including the weight of machineryCommunal kitchens, laundries 3.0 4.5Dining rooms, lounges, billiard rooms 2.0 2.7Toilet rooms 2.0 –Bedrooms, dormitories 1.5 1.8Corridors, hallways, stairs, landings, footbridges, etc. 3.0 4.5Balconies Same as rooms to which 1.5 per metre run

they give access but with concentrated ata minimum of 3.0 the outer edge

Cat walks – 1.0 at 1 m centres

Type 3. Hotels and motelsBoiler rooms, motor rooms, fan rooms and the like, 7.5 4.5

including the weight of machineryAssembly areas without fixed seating, dance halls 5.0 3.6Bars 5.0 –Assembly areas with fixed seatinga 4.0 –Corridors, hallways, stairs, landings, footbridges, etc. 4.0 4.5Kitchens, laundries 3.0 4.5Dining rooms, lounges, billiard rooms 2.0 2.7Bedrooms 2.0 1.8Toilet rooms 2.0 –Balconies Same as rooms to which 1.5 per metre run

they give access but with concentrated at thea minimum of 4.0 outer edge

Cat walks – 1.0 at 1 m centres

Note. a Fixed seating is seating where its removal and the use of the space for other purposes are improbable.

In most of the simple structures which will beconsidered in this book, the worst possible com-bination will arise due to the maximum dead andmaximum imposed loads acting on the structuretogether. In such cases, the partial safety factors fordead and imposed loads are 1.4 and 1.6 respect-ively (Fig. 2.3) and hence the design load is given by

Fig. 2.3

Design load = 1.4Gk + 1.6Q k

However, it should be appreciated that theoret-ically the design dead loads can vary between thecharacteristic and ultimate values, i.e. 1.0Gk and1.4Gk. Similarly, the design imposed loads canvary between zero and the ultimate value, i.e. 0.0Qk

and 1.6Qk. Thus for a simply supported beam withan overhang (Fig. 2.4(a)) the load cases shown inFigs 2.4(b)–(d) will need to be considered in orderto determine the design bending moments and shearforces in the beam.

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13

Fig. 2.4

Design loads acting on elements

Fig. 2.5 Typical beams and column support conditions.

commonly assumed support conditions at the endsof beams and columns respectively.

In design it is common to assume that all the jointsin the structure are pinned and that the sequence ofload transfer occurs in the order: ceiling/floor loads tobeams to columns to foundations to ground. Theseassumptions will considerably simplify calculationsand lead to conservative estimates of the designloads acting on individual elements of the struc-ture. The actual calculations to determine the forcesacting on the elements are best illustrated by anumber of worked examples as follows.

2.3 Design loads acting onelements

Once the design loads acting on the structure havebeen estimated it is then possible to calculate thedesign loads acting on individual elements. As waspointed out at the beginning of this chapter, thisusually requires the designer to make assumptionsregarding the support conditions and how the loadswill eventually be transmitted down to the ground.Figures 2.5(a) and (b) illustrate some of the more

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Example 2.2 Design loads on a floor beamA composite floor consisting of a 150 mm thick reinforced concrete slab supported on steel beams spanning 5 m andspaced at 3 m centres is to be designed to carry an imposed load of 3.5 kN m−2. Assuming that the unit mass of thesteel beams is 50 kg m−1 run, calculate the design loads on a typical internal beam.

UNIT WEIGHTS OF MATERIALSReinforced concreteFrom Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, theunit weight of reinforced concrete is

2400 × 10 = 24 000 N m−3 = 24 kN m−3

Steel beamsUnit mass of beam = 50 kg m−1 runUnit weight of beam = 50 × 10 = 500 N m−1 run = 0.5 kN m−1 run

LOADINGSlab

Slab dead load ( g k) = self-weight = 0.15 × 24 = 3.6 kN m−2

Slab imposed load (qk) = 3.5 kN m−2

Slab ultimate load = 1.4g k + 1.6q k = 1.4 × 3.6 + 1.6 × 3.5= 10.64 kN m−2

BeamBeam dead load ( gk) = self-weight = 0.5 kN m−1 runBeam ultimate load = 1.4g k = 1.4 × 0.5 = 0.7 kN m−1 run

DESIGN LOADEach internal beam supports a uniformly distributed load from a 3 m width of slab (hatched \\\\\\\) plus self-weight.Hence

Design load on beam = slab load + self-weight of beam

= 10.64 × 5 × 3 + 0.7 × 5

= 159.6 + 3.5 = 163.1 kN

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Design loads acting on elements

Example 2.3 Design loads on floor beams and columnsThe floor shown below with an overall depth of 225 mm is to be designed to carry an imposed load of 3 kN m−2 plusfloor finishes and ceiling loads of 1 kN m−2. Calculate the design loads acting on beams B1–C1, B2–C2 and B1–B3 andcolumns B1 and C1. Assume that all the column heights are 3 m and that the beam and column weights are 70 and60 kg m−1 run respectively.

UNIT WEIGHTS OF MATERIALSReinforced concreteFrom Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, theunit weight of reinforced concrete is

2400 × 10 = 24 000 N m−3 = 24 kN m−3

Steel beamsUnit mass of beam = 70 kg m−1 runUnit weight of beam = 70 × 10 = 700 N m−1 run = 0.7 kN m−1 run

Steel columnsUnit mass of column = 60 kg m−1 runUnit weight of column = 60 × 10 = 600 N m−1 run = 0.6 kN m−1 run

LOADINGSlab

Slab dead load (g k) = self-weight + finishes= 0.225 × 24 + 1 = 6.4 kN m−2

Slab imposed load (q k) = 3 kN m−2

Slab ultimate load = 1.4g k + 1.6q k

= 1.4 × 6.4 + 1.6 × 3 = 13.76 kN m−2

BeamBeam dead load (g k) = self-weight = 0.7 kN m−1 runBeam ultimate load = 1.4g k = 1.4 × 0.7 = 0.98 kN m−1 run

ColumnColumn dead load (g k) = 0.6 kN m−1 runColumn ultimate load = 1.4g k = 1.4 × 0.6 = 0.84 kN m−1 run

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DESIGN LOADS

Beam B1–C1Assuming that the slab is simply supported, beam B1–C1 supports a uniformly distributed load from a 1.5 m width ofslab (hatched //////) plus self-weight of beam. Hence

Design load on beam B1–C1 = slab load + self-weight of beam

= 13.76 × 6 × 1.5 + 0.98 × 6

= 123.84 + 5.88 = 129.72 kN

Since the beam is symmetrically loaded, RB2 and RC2 are the same and equal to 253.56/2 = 126.78 kN.

Beam B1–B3Assuming that the slab is simply supported, beam B1–B3 supports a uniformly distributed load from a 1.5 m width ofslab (shown cross-hatched) plus the self-weight of the beam and the reaction transmitted from beam B2–C2 whichacts as a point load at mid-span. Hence

Design load on beam B1–B3 = uniformly distributed load from slab plus self-weight of beam+ point load from reaction RB2

= (13.76 × 1.5 × 6 + 0.98 × 6) + 126.78

= 129.72 + 126.78 = 256.5 kN

Since the beam is symmetrically loaded,

RB1 = RC1 = 129.72/2 = 64.86 kN

Beam B2–C2Assuming that the slab is simply supported, beam B2–C2 supports a uniformly distributed load from a 3 m width ofslab (hatched \\\\\) plus its self-weight. Hence

Design load on beam B2–C2 = slab load + self-weight of beam

= 13.76 × 6 × 3 + 0.98 × 6

= 247.68 + 5.88 = 253.56 kN

Example 2.3 continued

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Structural analysis

Since the beam is symmetrically loaded,

RB1 = RB3 = 256.5/2 = 128.25 kN

Column B1

2.4 Structural analysisThe design axial loads can be used directly to size col-umns. Column design will be discussed more fullyin section 2.5. However, before flexural memberssuch as beams can be sized, the design bendingmoments and shear forces must be evaluated.Such calculations can be performed by a variety ofmethods as noted below, depending upon the com-plexity of the loading and support conditions:

1. equilibrium equations2. formulae3. computer methods.

Hand calculations are suitable for analysingstatically determinate structures such as simply sup-ported beams and slabs (section 2.4.1). For variousstandard load cases, formulae for calculating themaximum bending moments, shear forces anddeflections are available which can be used to rapidly

Column C1 supports the reactions from beams B1–C1 and C1–C3 and its self-weight. From the above, the reaction atC1 due to beam B1–C1 is 64.86 kN. Beam C1–C3 supports the reactions from B2–C2 (= 126.78 kN) and its self-weight (= 0.98 × 6) = 5.88 kN. Hence the reaction at C1 is (126.78 + 5.88)/2 = 66.33 kN. Since the column heightis 3 m, self-weight = 0.84 × 3 = 2.52 kN. Hence

Design load on column C1 = 64.86 + 66.33 + 2.52 = 133.71 kN

Column B1 supports the reactions from beams A1–B1, B1–C1 and B1–B3 and its self-weight. From the above, thereaction at B1 due to beam B1–C1 is 64.86 kN and from beam B1–B3 is 128.25 kN. Beam A1–B1 supports only itsself-weight = 0.98 × 3 = 2.94 kN. Hence reaction at B1 due to A1–B1 is 2.94/2 = 1.47 kN. Since the column height is3 m, self-weight of column = 0.84 × 3 = 2.52 kN. Hence

Design load on column B1 = 64.86 + 128.25 + 1.47 + 2.52

= 197.1 kN

Column C1


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analyse beams, as will be discussed in section 2.4.2.Alternatively, the designer may resort to using vari-ous commercially available computer packages, e.g.SAND. Their use is not considered in this book.

2.4.1 EQUILIBRIUM EQUATIONSIt can be demonstrated that if a body is in equilib-rium under the action of a system of external forces,all parts of the body must also be in equilibrium.

This principle can be used to determine the bend-ing moments and shear forces along a beam. Theactual procedure simply involves making fictitious‘cuts’ at intervals along the beam and applying theequilibrium equations given below to the cut por-tions of the beam.

Σ moments (M) = 0 (2.1)

Σ vertical forces (V ) = 0 (2.2)

From equation 2.1, taking moments about Z gives

126.78 × 1 − 42.26 × 1 × 0.5 − Mx=1 = 0

Hence

Mx=1 = 105.65 kN m

From equation 2.2, summing the vertical forces gives

126.78 − 42.26 × 1 − Vx=1 = 0

Example 2.4 Design moments and shear forces in beams usingequilibrium equations

Calculate the design bending moments and shear forces in beams B2–C2 and B1–B3 of Example 2.3.

BEAM B2–C2

Let the longitudinal centroidal axis of the beam be the x axis and x = 0 at support B2.

x ===== 0By inspection,

Moment at x = 0 (Mx=0) = 0

Shear force at x = 0 (Vx=0) = RB2 = 126.78 kN

x ===== 1Assuming that the beam is cut 1 m from support B2, i.e. x = 1 m, the moments and shear forces acting on the cutportion of the beam will be those shown in the free body diagram below:

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Structural analysis


126.78 × 2 − 42.26 × 2 × 1 − Mx=2 = 0

Hence

Mx=2 = 169.04 kN m

From equation 2.2, summing the vertical forces gives

126.78 − 42.26 × 2 − Vx=2 = 0

Hence

Vx=2 = 42.26 kN

If this process is repeated for values of x equal to 3, 4, 5 and 6 m, the following values of the moments and shearforces in the beam will result:

x(m) 0 1 2 3 4 5 6

M(kN m) 0 105.65 169.04 190.17 169.04 105.65 0V(kN ) 126.78 84.52 42.26 0 −42.26 −84.52 −126.78

This information is better presented diagrammatically as shown below:

Example 2.4 continuedHence

Vx=1 = 84.52 kN

x ===== 2The free body diagram for the beam, assuming that it has been cut 2 m from support B2, is shown below:

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20

Hence, the design moment (M) is 190.17 kN m. Note that this occurs at mid-span and coincides with the point ofzero shear force. The design shear force (V ) is 126.78 kN and occurs at the supports.

BEAM B1–B3

In (a) it is assumed that the beam is cut immediately to the left of the point load. In (b) the beam is cut immediatelyto the right of the point load. In (a), from equation 2.2, the shear force to the left of the cut, Vx=3,L, is given by

128.25 − 64.86 − Vx=3,L = 0

Hence

Vx=3,L = 63.39 kN


128.25 × 3 − 64.86 × 3/2 − Mx=3 = 0

Hence

Mx=3 = 287.46 kN m

In order to determine the shear force at x = 3 the following two load cases need to be considered:

Again, let the longitudinal centroidal axis of the beam be the x axis of the beam and set x = 0 at support B1.The steps outlined earlier can be used to determine the bending moments and shear forces at x = 0, 1 and 2 m andsince the beam is symmetrically loaded and supported, these values of bending moment and shear forces will applyat x = 4, 5 and 6 m respectively.

The bending moment at x = 3 m can be calculated by considering all the loading immediately to the left of thepoint load as shown below:


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21

Structural analysis

Example 2.4 continuedIn (b), from equation 2.2, the shear force to the right of the cut, Vx=3,R, is given by

128.25 − 64.86 − 126.78 − Vx=3,R = 0

Hence

Vx=3,R = −63.39 kN

Summarising the results in tabular and graphical form gives

x(m) 0 1 2 3 4 5 6

M(kN m) 0 117.44 213.26 287.46 213.26 117.44 0V(kN) 128.25 106.63 85.01 63.39|−63.39 −85.01 −106.63 −128.25

Hence, the design moment for beam B1–B3 is 287.46 kN m and occurs at mid-span and the design shear force is128.25 kN and occurs at the supports.

2.4.2 FORMULAEAn alternative method of determining the designbending moments and shear forces in beams in-volves using the formulae quoted in Table 2.3. Thetable also includes formulae for calculating themaximum deflections for each load case. The for-mulae can be derived using a variety of methods

for analysing statically indeterminate structures, e.g.slope deflection and virtual work, and the inter-ested reader is referred to any standard work onthis subject for background information. There willbe many instances in practical design where theuse of standard formulae is not convenient andequilibrium methods are preferable.

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22

Table 2.3 Bending moments, shear forces and deflections forvarious standard load cases

Loading Maximum Maximum Maximumbending shearing deflectionmoment force

WLEI

3

8

WL2

W

W2

WLEI

3

48

WL4

231296

3WLEI

WL6

W2

11768

3WLEI

W2

WL8

5384

3WLEI

W2

WL8

WL W

WLEI

3

3

WLEI

3

60

W2

WL6

WLEI

3

192(at supportsand at midspan)

WL8

W2

WL12

W2

WLEI

3

384at supports

WL24

at midspan

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23

Structural analysis

This load case can be solved using the principle of superposition which can be stated in general terms as follows: ‘Theeffect of several actions taking place simultaneously can be reproduced exactly by adding the effects of each caseseparately.’ Thus, the loading on beam B1–B3 can be considered to be the sum of a uniformly distributed load (Wudl)of 129.72 kN and a point load (Wpl) at mid-span of 126.78 kN.

By inspection, maximum moment and maximum shear force occur at beam mid-span and supports respectively. FromTable 2.3, design moment, M, is given by

M

W l

. = =

×8

253 56 68

= 190.17 kN m

and design shear force, V, is given by

V

W

. . = = =

2253 56

2126 78 kN

BEAM B1–B3

By inspection, the maximum bending moment and shear force for both load cases occur at beam mid-span andsupports respectively. Thus, the design moment, M, is given by

M

W l W l

.

. . = + =

×+

×=udl pl kN m

8 4129 72 6

8126 78 6

4287 46

and the design shear force, V, is given by

V

W W

.

. . = + = + =udl pl kN

2 2129 72

2126 78

2128 25

Example 2.5 Design moments and shear forces in beams using formulaeRepeat Example 2.4 using the formulae given in Table 2.3.

BEAM B2–C2

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24

2.5 Beam designHaving calculated the design bending moment andshear force, all that now remains to be done is toassess the size and strength of beam required.Generally, the ultimate limit state of bending willbe critical for medium-span beams which are mod-erately loaded and shear for short-span beamswhich are heavily loaded. For long-span beams theserviceability limit state of deflection may well becritical. Irrespective of the actual critical limitstate, once a preliminary assessment of the sizeand strength of beam needed has been made, it mustbe checked for the remaining limit states that mayinfluence its long-term integrity.

The processes involved in such a selection willdepend on whether the construction material be-haves (i) elastically or (ii) plastically. If the materialis elastic, it obeys Hooke’s law, that is, the stressin the material due to the applied load is directlyproportional to its strain (Fig. 2.6) where

Stress (σ) =

forcearea

Fig. 2.7 Strain and stress in an elastic beam.

and Strain (ε) =

change in lengthoriginal length

The slope of the graph of stress vs. strain (Fig. 2.6)is therefore constant, and this gradient is normallyreferred to as the elastic or Young’s modulus andis denoted by the letter E. It is given by

Young’s modulus (E) =

stressstrain

Note that strain is dimensionless but that bothstress and Young’s modulus are usually expressedin N mm−2.

A material is said to be plastic if it strains withouta change in stress. Plasticine and clay are plasticmaterials but so is steel beyond its yield point(Fig. 2.6). As will be seen in Chapter 3, reinforcedconcrete design also assumes that the materialbehaves plastically.

The structural implications of elastic and plasticbehaviour are best illustrated by considering howbending is resisted by the simplest of beams – arectangular section b wide and d deep.

2.5.1 ELASTIC CRITERIAWhen any beam is subject to load it bends as shownin Fig. 2.7(a). The top half of the beam is putinto compression and the bottom half into tension.In the middle, there is neither tension nor com-pression. This axis is normally termed the neutralaxis.

If the beam is elastic and stress and strain aredirectly proportional for the material, the variationin strain and stress from the top to middle to bot-tom is linear (Figs 2.7(b) and (c)). The maximumstress in compression and tension is σy. The aver-age stress in compression and tension is σy/2. Hencethe compressive force, Fc, and tensile force, Ft,acting on the section are equal and are given byFig. 2.6 Stress–strain plot for steel.

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25

atively if b and d are known, the required materialstrength can be evaluated.

Equation 2.3 is more usually written as

M = σZ (2.5)

or

M =

σIy

(2.6)

where Z is the elastic section modulus and is equalto bd 2/6 for a rectangular beam, I is the moment ofinertia or, more correctly, the second moment ofarea of the section and y the distance from theneutral axis (Fig. 2.7(a)). The elastic section modu-lus can be regarded as an index of the strength ofthe beam in bending. The second moment of areaabout an axis x–x in the plane of the section isdefined by

Ixx = ∫ y2dA (2.7)

Second moments of area of some common shapesare given in Table 2.4.

2.5.2 PLASTIC CRITERIAWhile the above approach would be suitable fordesign involving the use of materials which have alinear elastic behaviour, materials such as reinforcedconcrete and steel have a substantial plastic per-formance. In practice this means that on reachingan elastic yield point the material continues to de-form but with little or no change in maximum stress.

Figure 2.8 shows what this means in terms ofstresses in the beam. As the loading on the beamincreases extreme fibre stresses reach the yield point,σy, and remain constant as the beam continues tobend. A zone of plastic yielding begins to penetrateinto the interior of the beam, until a point is reachedimmediately prior to complete failure, when practic-ally all the cross-section has yielded plastically. Theaverage stress at failure is σy, rather than σy/2 as was

Fc = Ft = F = stress × area

= =

σ σy y

2 2 4bd bd

The tensile and compression forces are separ-ated by a distance s whose value is equal to 2d /3(Fig. 2.7(a)). Together they make up a couple, or mo-ment, which acts in the opposite sense to the designmoment. The value of this moment of resistance,Mr, is given by

M r = Fs =

bd d bdσ σy y

423 6

2

= (2.3)

At equilibrium, the design moment in the beamwill equal the moment of resistance, i.e.

M = M r (2.4)

Provided that the yield strength of the material,i.e. σy is known, equations 2.3 and 2.4 can be usedto calculate suitable dimensions for the beamneeded to resist a particular design moment. Altern-

Fig. 2.8 Bending failure of a beam (a) below yield(elastic); ( b) at yield point (elastic); (c) beyond yield(partially plastic); (d) beyond yield ( fully plastic).

Beam design

I

DB dbyy = − 3 3

12

Table 2.4 Second moments of area

Shape Second moment of area

Ixx = bd3/12

Iyy = db3/12

Ixx = bh3/36Iaa = bh3/12

Ixx = πR4/4Ipolar = 2Ixx

I

bdaa =

3

3

I

BD bdxx =

−

3 3

12

I

BD bdxx =

−

3 3

12

I

TB dtyy = +

212 12

3 3

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26

combination of buckling followed by crushing ofthe material(s) (Fig. 1.3). These columns will tendto have lower load-carrying capacities, a fact whichis taken into account by reducing the design stressesin the column. The design stresses are related tothe slenderness ratio of the column, which is foundto be a function of the following factors:

1. geometric properties of the column cross-section, e.g. lateral dimension of column, radiusof gyration;

2. length of column;3. support conditions (Fig. 2.5).

The radius of gyration, r, is a sectional propertywhich provides a measure of the column’s abilityto resist buckling. It is given by

r = (I/A)1/2

(2.11)

Generally, the higher the slenderness ratio, thegreater the tendency for buckling and hence thelower the load capacity of the column. Most prac-tical reinforced concrete columns are designed tofail by crushing and the design equations for thismedium are based on equation 2.10 (Chapter 3).Steel columns, on the other hand, are designed tofail by a combination of buckling and crushing.Empirical relations have been derived to predictthe design stress in steel columns in terms of theslenderness ratio (Chapter 4).

found to be the case when the material was assumedto have a linear-elastic behaviour. The moment ofresistance assuming plastic behaviour is given by

M Fs

bd d bdSr = = = =

σ σσy y

y2 42

2

(2.8)

where S is the plastic section modulus and isequal to bd 2/4 for rectangular beams. By setting thedesign moment equal to the moment of resistanceof the beam its size and strength can be calculatedaccording to plastic criteria (Example 2.6).

2.6 Column designGenerally, column design is relatively straightfor-ward. The design process simply involves makingsure that the design load does not exceed the loadcapacity of the column, i.e.

Load capacity ≥ design axial load (2.9)

Where the column is required to resist a predomin-antly axial load, its load capacity is given by

Load capacity = design stress × area ofcolumn cross-section (2.10)

The design stress is related to the crushingstrength of the material(s) concerned. However, notall columns fail in this mode. Some fail due to a

Example 2.6 Elastic and plastic moments of resistance of abeam section

Calculate the moment of resistance of a beam 50 mm wide by 100 mm deep with σy = 20 N mm−2 according to (i)elastic criteria and (ii) plastic criteria.

ELASTIC CRITERIAFrom equation 2.3,

M

bdr,el y N mm . = = × ×

= ×2 2

6

650 100 20

61 67 10σ

PLASTIC CRITERIAFrom equation 2.8,

M

bdr,pl

y

4N mm

. = =

× ×= ×

2 2650 100 20

42 5 10

σ

Hence it can be seen that the plastic moment of resistance of the section is greater than the maximum elasticmoment of resistance. This will always be the case but the actual difference between the two moments will dependupon the shape of the section.

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27

Summary

In most real situations, however, the loads willbe applied eccentric to the axes of the column.Columns will therefore be required to resist axialloads and bending moments about one or bothaxes. In some cases, the bending moments may besmall and can be accounted for in design simplyby reducing the allowable design stresses in the ma-terial(s). Where the moments are large, for instancecolumns along the outside edges of buildings,the analysis procedures become more complex andthe designer may have to resort to the use of designcharts. However, as different codes of practice treatthis in different ways, the details will be discussedseparately in the chapters which follow.

2.7 SummaryThis chapter has presented a simple but conserv-ative method for assessing the design loads actingon individual members of building structures. Thisassumes that all the joints in the structure arepinended and that the sequence of load transferoccurs in the order: ceiling /floor loads to beams tocolumns to foundations to ground. The design loadsare used to calculate the design bending moments,shear forces, axial loads and deflections experiencedby members. In combination with design strengthsand other properties of the construction medium,the sizes of the structural members are determined.

Example 2.7 Analysis of column sectionDetermine whether the reinforced concrete column cross-section shown below would be suitable to resist an axialload of 1500 kN. Assume that the design compressive strengths of the concrete and steel reinforcement are 14 and375 N mm−2 respectively.

Area of steel bars = 4 × (π202/4) = 1256 mm2

Net area of concrete = 300 × 300 − 1256 = 88 744 mm2

Load capacity of column = force in concrete + force in steel

= 14 × 88 744 + 375 × 1256

= 1 713 416 N = 1713.4 kN

Hence the column cross-section would be suitable to resist the design load of 1500 kN.

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28

D = 360 mm x x

t = 7 mm

T = 12 mmB = 170 mm

Questions1. For the beams shown, calculate and

sketch the bending moment and shearforce diagrams.

2. For the three load cases shown inFig. 2.4, sketch the bending moment andshear force diagrams and hence determinethe design bending moments and shearforces for the beam. Assume the mainspan is 6 m and the overhang is 2 m. Thecharacteristic dead and imposed loads arerespectively 20 kNm−1 and 10 kNm−1.

3. (a) Calculate the area and the major axismoment of inertia, elastic modulus,plastic modulus and the radius ofgyration of the steel I-section shownbelow.

according to (i) elastic and (ii) plasticcriteria. Use your results to determinethe working and collapse loads of abeam with this cross-section, 6 m longand simply supported at its ends.Assume the loading on the beam isuniformly distributed.

4. What are the most common ways inwhich columns can fail? List and discussthe factors that influence the load carryingcapacity of columns.

5. The water tank shown in Fig. 2.9 issubjected to the following characteristicdead (Gk), imposed (Q k) and wind loads(Wk) respectively(i) 200 kN, 100 kN and 50 kN (Load

case 1)(ii) 200 kN, 100 kN and 75 kN (Load

case 2).Assuming the support legs are pinned atthe base, determine the design axial forcesin both legs by considering the followingload combinations:(a) dead plus imposed(b) dead plus wind(c) dead plus imposed plus wind.Refer to Table 3.4 for relevant loadfactors.

(b) Assuming the design strength of steel,σy, is 275 Nmm−2, calculate themoment of resistance of the I-section Fig. 2.9

Gk,Qk1.2 m

5 m

Leg R

Leg L

2 m

Wk

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29

PART TWO

STRUCTURALDESIGN TO BRITISHSTANDARDS

Part One has discussed the principles of limit statedesign and outlined general approaches towardsassessing the sizes of beams and columns in build-ing structures. Since limit state design forms thebasis of the design methods in most modern codesof practice on structural design, there is consider-able overlap in the design procedures presented inthese codes.

The aim of this part of the book is to givedetailed guidance on the design of a number ofstructural elements in the four media: concrete,steel, masonry and timber to the current BritishStandards. The work has been divided into fourchapters as follows:

1. Chapter 3 discusses the design procedures inBS 8110: Structural use of concrete relating tobeams, slabs, pad foundations, retaining wallsand columns.

2. Chapter 4 discusses the design procedures in BS5950: Structural use of steelwork in buildings relat-ing to beams, columns, floors and connections.

3. Chapter 5 discusses the design procedures inBS 5628: Code of practice for use of masonry relat-ing to unreinforced loadbearing and panel walls.

4. Chapter 6 discusses the design procedures in BS5268: Structural use of timber relating to beams,columns and stud walling.

9780415467193_C03a 9/3/09, 12:41 PM29

Design in reinforced concrete to BS 8110

30

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31

Chapter 3

Design in reinforcedconcrete to BS 8110

the recommendations given in various documentsincluding BS 5400: Part 4: Code of practice fordesign of concrete bridges, BS 8007: Code of prac-tice for the design of concrete structures for retainingaqueous liquids and BS 8110: Structural use of con-crete. Since the primary aim of this book is to giveguidance on the design of structural elements, thisis best illustrated by considering the contents ofBS 8110.

BS 8110 is divided into the following three parts:

Part 1: Code of practice for design and construction.Part 2: Code of practice for special circumstances.Part 3: Design charts for singly reinforced beams, doubly

reinforced beams and rectangular columns.

Part 1 covers most of the material required foreveryday design. Since most of this chapter isconcerned with the contents of Part 1, it shouldbe assumed that all references to BS 8110 refer toPart 1 exclusively. Part 2 covers subjects such astorsional resistance, calculation of deflections andestimation of crack widths. These aspects of designare beyond the scope of this book and Part 2, there-fore, is not discussed here. Part 3 of BS 8110 con-tains charts for use in the design of singly reinforcedbeams, doubly reinforced beams and rectangularcolumns. A number of design examples illustratingthe use of these charts are included in the relevantsections of this chapter.

3.2 Objectives and scopeAll reinforced concrete building structures arecomposed of various categories of elements includ-ing slabs, beams, columns, walls and foundations(Fig. 3.1). Within each category is a range of ele-ment types. The aim of this chapter is to describethe element types and, for selected elements, togive guidance on their design.

This chapter is concerned with the detailed design ofreinforced concrete elements to British Standard 8110.A general discussion of the different types of commonlyoccurring beams, slabs, walls, foundations and col-umns is given together with a number of fully workedexamples covering the design of the following elements:singly and doubly reinforced beams, continuous beams,one-way and two-way spanning solid slabs, pad founda-tion, cantilever retaining wall and short braced columnssupporting axial loads and uni-axial or bi-axial bend-ing. The section which deals with singly reinforced beamsis, perhaps, the most important since it introduces thedesign procedures and equations which are common tothe design of the other elements mentioned above, withthe possible exception of columns.

3.1 IntroductionReinforced concrete is one of the principal materialsused in structural design. It is a composite material,consisting of steel reinforcing bars embedded inconcrete. These two materials have complementaryproperties. Concrete, on the one hand, has highcompressive strength but low tensile strength. Steelbars, on the other, can resist high tensile stressesbut will buckle when subjected to comparativelylow compressive stresses. Steel is much moreexpensive than concrete. By providing steel barspredominantly in those zones within a concretemember which will be subjected to tensile stresses,an economical structural material can be producedwhich is both strong in compression and strongin tension. In addition, the concrete provides cor-rosion protection and fire resistance to the morevulnerable embedded steel reinforcing bars.

Reinforced concrete is used in many civilengineering applications such as the constructionof structural frames, foundations, retaining walls,water retaining structures, highways and bridges.They are normally designed in accordance with

9780415467193_C03a 9/3/09, 12:41 PM31


32

3. material properties4. loading5. stress–strain relationships6. durability and fire resistance.

The detailed design of beams, slabs, foundations,retaining walls and columns will be discussed insections 3.9, 3.10, 3.11, 3.12 and 3.13, respectively.

3.3 SymbolsFor the purpose of this book, the following sym-bols have been used. These have largely been takenfrom BS 8110. Note that in one or two cases thesame symbol is differently defined. Where thisoccurs the reader should use the definition mostappropriate to the element being designed.

Geometric properties:

b width of sectiond effective depth of the tension reinforcementh overall depth of sectionx depth to neutral axisz lever armd ′ depth to the compression reinforcementb effective spanc nominal cover to reinforcement

Bending:

Fk characteristic loadgk, Gk characteristic dead loadqk, Qk characteristic imposed loadwk, Wk characteristic wind loadfk characteristic strengthfcu characteristic compressive cube strength

of concretefy characteristic tensile strength of

reinforcementγ f partial safety factor for loadγm partial safety factor for material

strengthsK coefficient given by M/fcubd 2

K ′ coefficient given by Mu/fcubd 2 = 0.156 whenredistribution does not exceed 10 per cent

M design ultimate momentMu design ultimate moment of resistanceAs area of tension reinforcementAs′ area of compression reinforcementΦ diameter of main steelΦ′ diameter of links

Shear:

fyv characteristic strength of linkssv spacing of links along the member

Fig. 3.1 Some elements of a structure.

Fig. 3.2 Cantilever retaining wall.

A great deal of emphasis has been placed in thetext to highlight the similarities in structural beha-viour and, hence, design of the various categories ofelements. Thus, certain slabs can be regarded fordesign purposes as a series of transversely connectedbeams. Columns may support slabs and beamsbut columns may also be supported by (groundbearing) slabs and beams, in which case the latterare more commonly referred to as foundations.Cantilever retaining walls are usually designed asif they consist of three cantilever beams as shownin Fig. 3.2. Columns are different in that they areprimarily compression members rather than beamsand slabs which predominantly resist bending.Therefore columns are dealt with separately at theend of the chapter.

Irrespective of the element being designed, thedesigner will need a basic understanding of the fol-lowing aspects which are discussed next:

1. symbols2. basis of design

Deflectedshape ofwall

Wall

Horizontalpressure

Pressure

Deflected shapeat base

Roof

2nd Floor

1st Floor

Beams

Floor slabs

Columns

Walls

Foundations

9780415467193_C03a 9/3/09, 12:41 PM32

33

V design shear force due to ultimateloads

v design shear stressvc design concrete shear stressA sv total cross-sectional area of shear

reinforcement

Compression:

b width of columnh depth of columnbo clear height between end restraintsbe effective heightbex effective height in respect of x-x axisbey effective height in respect of y-y axisN design ultimate axial loadAc net cross-sectional area of concrete in

a columnAsc area of longitudinal reinforcement

3.4 Basis of designThe design of reinforced concrete elements toBS 8110 is based on the limit state method. Asdiscussed in Chapter 1, the two principal categoriesof limit states normally considered in design are:

(i) ultimate limit state(ii) serviceability limit state.

The ultimate limit state models the behaviourof the element at failure due to a variety of mech-anisms including excessive bending, shear andcompression or tension. The serviceability limit statemodels the behaviour of the member at workingloads and in the context of reinforced concretedesign is principally concerned with the limit statesof deflection and cracking.

Having identified the relevant limit states, thedesign process simply involves basing the designon the most critical one and then checking for theremaining limit states. This requires an understand-ing of

1. material properties2. loadings.

3.5 Material propertiesThe two materials whose properties must be knownare concrete and steel reinforcement. In the caseof concrete, the property with which the designeris primarily concerned is its compressive strength. Fig. 3.3 Normal frequency distribution of strengths.

Meanstrength

Num

ber

of r

esul

ts

5% ofresults

fm

1.64 s.d.

Strengthfcu

Material properties

For steel, however, it is its tensile strength capacitywhich is important.

3.5.1 CHARACTERISTIC COMPRESSIVESTRENGTH OF CONCRETE, fcu

Concrete is a mixture of water, coarse and fineaggregate and a cementitious binder (normally Port-land cement) which hardens to a stone like mass.As can be appreciated, it is difficult to produce ahomogeneous material from these components.Furthermore, its strength and other properties mayvary considerably due to operations such as trans-portation, compaction and curing.

The compressive strength of concrete is usuallydetermined by carrying out compression tests on28-day-old, 100 mm cubes which have been pre-pared using a standard procedure laid down in BSEN 12390-1 (2000). An alternative approach is touse 100 mm diameter by 200 mm long cylinders.Irrespective of the shape of the test specimen, ifa large number of compression tests were carriedout on samples made from the same mix it wouldbe found that a plot of crushing strength againstfrequency of occurrence would approximate to anormal distribution (Fig. 3.3).

For design purposes it is necessary to assume aunique value for the strength of the mix. However,choosing too high a value will result in a high prob-ability that most of the structure will be constructedwith concrete having a strength below this value.Conversely, too low a value will result in inefficientuse of the material. As a compromise betweeneconomy and safety, BS 8110 refers to the charac-teristic strength ( fcu) which is defined as the valuebelow which not more than 5 per cent of the testresults fall.

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34

Table 3.2 Strength of reinforcement(Table 3.1, BS 8110)

Reinforcement type Characteristic strength, fy

(Nmm−2)

Hot rolled mild steel 250High-yield steel (hot rolled 500

or cold worked)

Table 3.1 Concrete compressive strength classes

Concrete Designated Characteristic cubestrength classes concrete strength, fcu (Nmm−2)

C 20/25 RC 20/25 25C 25/30 RC 25/30 30C 28/35 RC 28/35 35C 32/40 RC 32/40 40C 35/45 RC 35/45 45C 40/50 RC 40/50 50C 50/60 – 60

3.5.2 CHARACTERISTIC STRENGTH OFREINFORCEMENT, fy

Concrete is strong in compression but weak intension. Because of this it is normal practice toprovide steel reinforcement in those areas wheretensile stresses in the concrete are most likely to de-velop. Consequently, it is the tensile strength of thereinforcement which most concerns the designer.

The tensile strength of steel reinforcement canbe determined using the procedure laid down inBS EN 10002: Part 1. The tensile strength willalso vary ‘normally’ with specimens of the samecomposition. Using the same reasoning as above,BS 8110 recommends that design should be basedon the characteristic strength of the reinforcement( fy) and gives typical values for mild steel and high-yield steel reinforcement, the two reinforcementtypes available in the UK, of 250 Nmm−2 and 500Nmm−2 respectively (Table 3.2). High-yield rein-forcement is mostly used in practice nowadays.

3.5.3 DESIGN STRENGTHTests to determine the characteristic strengths ofconcrete and steel reinforcement are carried out onnear perfect specimens, which have been preparedunder laboratory conditions. Such conditions willseldom exist in practice. Therefore it is undesirableto use characteristic strengths to size members.

To take account of differences between actualand laboratory values, local weaknesses and inac-curacies in assessment of the resistances of sections,the characteristic strengths ( fk) are divided byappropriate partial safety factor for strengths (γm),obtained from Table 3.3. The resulting values aretermed design strengths and it is the design strengthswhich are used to size members.

Design strength =

fk

mγ (3.1)

It should be noted that for the ultimate limitstate the partial safety factor for reinforcement (γms)is always 1.15, but for concrete (γmc) assumes

The characteristic and mean strength ( fm) of asample are related by the expression:

fcu = fm − 1.64 s.d.

where s.d. is the standard deviation. Thus assumingthat the mean strength is 35 Nmm−2 and standarddeviation is 3 Nmm−2, the characteristic strength ofthe mix is 35 − 1.64 × 3 = 30 Nmm−2.

The characteristic compressive strength of con-crete can be identified by its ‘strength class’. Table3.1 shows typical compressive strength classes ofconcrete commonly used in reinforced concrete de-sign. Note that the strength class consists of thecharacteristic cylinder strength of the mix followedby its characteristic cube strength. For example, aclass C25/30 concrete has a characteristic cylinderstrength of 25 Nmm−2 and a characteristic cubestrength of 30 Nmm−2. Nevertheless, like previouseditions of BS 8110, the design rules in the latestedition are based on characteristic cube not cylin-der strengths. In general, concrete strength classesin the range C20/25 and C50/60 can be designedusing BS 8110.

Table 3.1 also shows the two common approachesto the specification of concrete recommended inBS 8500, namely designed and designated. In manyapplications the most straightforward approachis to use a designated concrete which simply in-volves specifying the strength class, e.g. RC 20/25,and the maximum aggregate size. However, thisapproach may not be suitable for foundations,for example if ground investigations indicate theconcrete will be exposed to an aggressive chemicalenvironment. Under these circumstances a designedmix may be required and the designer will needto specify not only the strength class, i.e. C20/25,and the maximum aggregate size but also themaximum permissible water/cement ratio, minimumcement content, permitted cement or combinationtypes, amongst other aspects.

9780415467193_C03a 9/3/09, 12:41 PM34

35

different values depending upon the stress typeunder consideration. Furthermore, the partial safetyfactors for concrete are all greater than that for rein-forcement since concrete quality is less controllable.

3.6 LoadingIn addition to the material properties, the designerneeds to know the type and magnitude of the load-ing to which the structure may be subject duringits design life.

The loads acting on a structure are dividedinto three basic types: dead, imposed and wind(section 2.2). Associated with each type of loadingthere are characteristic and design values whichmust be assessed before the individual elements ofthe structure can be designed. These aspects arediscussed next.

3.6.1 CHARACTERISTIC LOADAs noted in Chapter 2, it is not possible to applystatistical principles to determine characteristic dead(Gk), imposed (Qk) and wind (Wk) loads simplybecause there are insufficient data. Therefore, thecharacteristic loads are taken to be those given inthe following documents:

Table 3.4 Values of γ f for various load combinations (based on Table 2.1, BS 8110)

Load combination Load type

Dead, Gk Imposed, Q k Wind, Wk

Adverse Beneficial Adverse Beneficial

1. Dead and imposed 1.4 1.0 1.6 0 –2. Dead and wind 1.4 1.0 – – 1.43. Dead and wind and imposed 1.2 1.2 1.2 1.2 1.2

Table 3.3 Values of γm for the ultimate limitstate (Table 2.2, BS 8110)

Material/Stress type Partial safetyfactor, γm

Reinforcement 1.15Concrete in flexure or axial load 1.50Concrete shear strength without shear 1.25

reinforcementConcrete bond strength 1.40Concrete, others (e.g. bearing stress) ≥ 1.50

Loading

1. BS 648: Schedule of weights for building materials.2. BS 6399: Design loadings for buildings, Part 1:

Code of practice for dead and imposed loads; Part2: Code of practice for wind loads; Part 3: Code ofpractice for imposed roof loads

3.6.2 DESIGN LOADVariations in the characteristic loads may arisedue to a number of reasons such as errors in theanalysis and design of the structure, constructionalinaccuracies and possible unusual load increases.In order to take account of these effects, the char-acteristic loads (Fk) are multiplied by the appropri-ate partial safety factor for loads (γf), taken fromTable 3.4, to give the design loads acting on thestructure:

Design load = γfFk (3.2)

Generally, the ‘adverse’ factors will be used toderive the design loads acting on the structure. Forexample, for single-span beams subject to only deadand imposed loads the appropriate values of γf aregenerally 1.4 and 1.6 respectively (Fig. 3.4(a)). How-ever, for continuous beams, load cases must be ana-lysed which should include maximum and minimumdesign loads on alternate spans (Fig. 3.4(b)).

The design loads are used to calculate thedistribution of bending moments and shear forcesin the structure usually using elastic analysis meth-ods as discussed in Chapter 2. At no point shouldthey exceed the corresponding design strengths ofthe member, otherwise failure of the structure mayarise.

The design strength is a function of the distribu-tion of stresses in the member. Thus, for the simplecase of a steel bar in direct tension the designstrength is equal to the cross-sectional area of thebar multiplied by the average stress at failure(Fig 3.5). The distribution of stresses in reinforcedconcrete members is usually more complicated, butcan be estimated once the stress–strain behaviour

9780415467193_C03a 9/3/09, 12:41 PM35


36

1.4gk + 1.6qk

(a)

(b)

1.4gk + 1.6qk

1.4gk + 1.6qk 1.4gk + 1.6qk1.0gk 1.0gk

of the concrete and steel reinforcement is known.This aspect is discussed next.

3.7 Stress-strain curves3.7.1 STRESS-STRAIN CURVE FOR CONCRETEFigure 3.6 shows a typical stress–strain curve fora concrete cylinder under uniaxial compression.Note that the stress–strain behaviour is never trulylinear and that the maximum compressive stress

Fig. 3.5 Design strength of the bar. Design strength,T, = σ.A, where σ is the average stress at failure andA the cross-sectional area of the bar.

σT T

A

at failure is approximately 0.8 × characteristic cubestrength (i.e. 0.8fcu).

However, the actual behaviour is rather com-plicated to model mathematically and, therefore,BS 8110 uses the modified stress–strain curveshown in Fig. 3.7 for design. This assumes that thepeak stress is only 0.67 (rather than 0.8) times thecharacteristic strength and hence the design stressfor concrete is given by

Design compressive

= ≈

. .

0 670 45

ffcu

mccuγstress for concrete (3.3)

In other words, the failure stress assumed in de-sign is approximately 0.45/0.8 = 56 per cent of theactual stress at failure when near perfect specimensare tested.

3.7.2 STRESS–STRAIN CURVE FOR STEELREINFORCEMENT

A typical tensile stress–strain curve for steel rein-forcement is shown in Fig. 3.8. It can be divided

0.67f cu

γm Parabolic curve

Stress

5.5fcu

γmkN/mm2

2.4 × 10−4 (f cu/γm)

Strain 0.0035

Fig. 3.7 Design stress–strain curve for concrete incompression (Fig. 2.1, BS 8110).

Fig. 3.6 Actual stress–strain curve for concrete incompression.

StressPeak stress ≈ 0.8fcu

Strain

Fig. 3.4 Ultimate design loads: (a) single span beam; (b) continuous beam.

9780415467193_C03a 9/3/09, 12:41 PM36

37

Fig. 3.8 Actual stress–strain curve for reinforcement.

Stress

f y

Strain

Fig. 3.9 Design stress–strain curve for reinforcement(Fig. 2.2, BS 8110).

Compressionfy /γm

fy /γmTension

200 kN/mm2

Strain

Str

ess

Durability and fire resistance

respect of durability and fire resistance since theserequirements are common to several of the ele-ments which will be subsequently discussed.

3.8 Durability and fire resistanceApart from the need to ensure that the design isstructurally sound, the designer must also verifythe proper performance of the structure in service.Principally this involves consideration of the twolimit states of (i) durability and (ii) fire resistance.It should be noted that much of the detailed guid-ance on durability design is given in BS 8500-1not BS 8110.

3.8.1 DURABILITYMany concrete structures are showing signs ofsevere deterioration after only a few years of ser-vice. Repair of these structures is both difficult andextremely costly. Therefore, over recent years, mucheffort has been directed towards improving thedurability requirements, particularly with regard tothe protection of steel reinforcement in concretefrom corrosion caused by carbonation and chlorideattack (Table 3.5). The other main mechanismsof concrete deterioration which are addressed inBS 8500-1 are freeze/thaw attack, sulphate attackand alkali/silica reaction.

In general, the durability of concrete structuresis largely achieved by imposing limits on:

1. the minimum strength class of concrete;2. the minimum cover to reinforcement;3. the minimum cement content;4. the maximum water/cement ratio;5. the cement type or combination;6. the maximum allowable surface crack width.

Other measures may include the specification ofparticular types of admixtures, restrictions on theuse of certain types of aggregates, the use of detailsthat ensure concrete surfaces are free draining andgood workmanship.

Generally speaking, the risk of freeze/thawattack and reinforcement corrosion decreases withincreasing compressive strength of concrete. In thecase of freeze/thaw attack this is largely becauseof the concomitant increase in tensile capacity ofthe concrete, which reduces the risk of crackingand spalling when water in the concrete expandson freezing. The use of an air entraining agent alsoenhances the frost resistance of concrete and is awell-established method of achieving this require-ment in practice.

into two regions: (i) an elastic region where strainis proportional to stress and (ii) a plastic regionwhere small increases in stress produce largeincreases in strain. The change from elastic to plasticbehaviour occurs at the yield stress and is signi-ficant since it defines the characteristic strength ofreinforcement ( fy).

Once again, the actual material behaviour israther complicated to model mathematically andtherefore BS 8110 modifies it to the form shown inFig. 3.9 which also includes the idealised stress–strain relationship for reinforcement in compression.The maximum design stress for reinforcement intension and compression is given by

Design stress for reinforcement =

fy

msγ (3.4)

From the foregoing it is possible to determinethe distribution of stresses at a section and hencecalculate the design strength of the member. Thelatter is normally carried out using the equationsgiven in BS 8110. However, before consideringthese in detail, it is useful to pause for a momentin order to introduce BS 8110’s requirements in

9780415467193_C03a 9/3/09, 12:41 PM37


38

Table 3.5 Exposure classes related to environmental conditions in accordance with BS EN 206 andBS 8500

Class Description of the environment Informative examples where exposure classes may occur

1. No risk of corrosionX0 For concrete with reinforcement Concrete inside buildings with very low (around 35%) humidity

or embedded metal: very dry

2. Corrosion induced by carbonationXC1 Dry or permanently wet Concrete inside building with low air humidity

Concrete permanently submerged in waterXC2 Wet, rarely dry Concrete surfaces subject to long-term water contact; many

foundationsXC3 Moderate humidity Concrete inside buildings with moderate or high humidity

External concrete sheltered from rainXC4 Cyclic wet and dry Concrete surfaces subject to water contact, not within exposure class

XC2

3. Corrosion induced by chloridesXD1 Moderate humidity Concrete exposed to airborne chloridesXD2 Wet, rarely dry Concrete totally immersed in water containing chlorides, e.g.

swimming poolsConcrete exposed to industrial waters containing chlorides

XD3 Cyclic wet and dry Parts of bridges exposed to spray containing chloridesPavements, car park slabs

4. Corrosion induced by chlorides from sea waterXS1 Exposed to air borne salt but not Structures near to or on the coast in direct contact with sea waterXS2 Permanently submerged Parts of marine structuresXS3 Tidal, splash and spray zones Parts of marine structures

5. Freeze/thaw attackXF1 Moderate water saturation, Vertical concrete surfaces exposed to rain and freezing

without deicing agentXF2 Moderate water saturation, with Vertical concrete surfaces of road structures exposed to freezing and

deicing agent airborne deicing agentsXF3 High water saturation, without Horizontal surfaces exposed to rain and freezing

deicing agentXF4 Moderate water saturation, with Road and bridge decks exposed to deicing agents; concrete surfaces

deicing agent exposed to direct spray containing deicing agents and freezing;splash zone of marine structures exposed to freezing

6. Chemical attackACEC See Table 3.8 Reinforced concrete in contact with the ground, e.g. many

foundations

The reduction in risk of reinforcement corrosionwith increasing compressive strength of concrete islinked to the associated reduction in the permeabil-ity of concrete. A low permeability mix enhancesdurability by reducing the rate of carbonationand chloride penetration into concrete as well as

restricting ionic movement within the concreteduring corrosion. These features not only increasethe time to the onset of corrosion (initiation time)but also reduce the subsequent rate of corrosionpropagation. The permeability of concrete is alsoinfluenced by water/cement ratio, cement content

9780415467193_C03a 9/3/09, 12:41 PM38

39


as well as type/composition of the cement, e.g.CEM I, IIB-V, IIIA, IIIB, IVB (see key at bottomof Table 3.6 for details), which provide addi-tional means of enhancing concrete durability. It isnoteworthy that the link between carbonation-induced corrosion and concrete permeability is lesspronounced than for chloride-induced corrosion.This is reflected in the values of nominal cover toreinforcement for carbonation-induced corrosionwhich are largely independent of cement type(Table 3.6). Nevertheless, for both carbonationand chloride attack a good thickness of concretecover is vital for corrosion protection as is the needto limit crack widths, in particular where cracksfollow the line of the reinforcement (coincidentcracks).

Sulphate attack is normally countered by speci-fying sulphate resisting Portland cement (SRPC).The risk of alkali-silica reaction can be reduced byspecifying non-reactive aggregate and/or cementi-tious materials with a low alkali content. It shouldbe noted that deterioration of concrete is rarelydue to a single cause, which can sometimes makespecification tricky.

Table 3.5 (taken from BS 8500-1) shows the rangeof exposure classes relevant to concrete construc-tion. As can be seen, the exposure classes are gen-erally broken down into the major concretedeterioration processes discussed above. Althoughthis system allows for the possibility of no risk ofcorrosion, i.e. exposure class X0, it is recommendedthat it is not applied to reinforced concrete as

Table 3.6 Concrete quality and cover to reinforcement for durability for an intended working life ofat least 50 years (based on Table A4 BS 8500-1)

Class Cement Strength class, max. w/c ratio, min. cement or combination content (kg/m3) orcombination type1 equivalent designated concrete

Nominal cover to reinforcement15+∆c 20+∆c 25+∆c 30+∆c 35+∆c 40+∆c 45+∆c 50+∆c

1. No risk of corrosionX0 All Not recommended for reinforced concrete structures2. Corrosion induced by carbonationXC1 All C20/25,

0.70, use the same grade of concrete240

XC2 All – – C25/30,0.65,260

XC3/ All except IVB – C40/50, C30/37, C28/35, C25/30,XC4 0.45, 0.55, 0.60, 0.65,

340 300 280 2603. Corrosion induced by chloridesXD1 All – – C40/50, C32/40, C28/35,

0.45, 0.55, 0.60,360 320 300

XD2 CEM I, IIA, IIB–S, SRPC – – – C40/50, C32/40, C28/35,0.40, 0.50, 0.55,380 340 320

IIB–V, IIIA – – – C35/45, C28/35, C25/30,0.40, 0.50, 0.55,380 340 320

IIIB–V, IVB – – – C32/40, C25/30, C20/25,0.40, 0.50, 0.55,380 340 320

9780415467193_C03a 9/3/09, 12:41 PM39


40

Table 3.6 (cont’d )

Class Cement Strength class, max. w/c ratio, min. cement or combination content (kg/m3) orcombination type1 equivalent designated concrete

Nominal cover to reinforcement15+∆c 20+∆c 25+∆c 30+∆c 35+∆c 40+∆c 45+∆c 50+∆c

XD3 CEM I, IIA, IIB–S, SRPC – – – – – C45/55, C40/50, C35/45,0.35, 0.40, 0.45,380 380 360

IIB–V, IIIA – – – – – C35/45, C32/40, C28/35,0.40, 0.45, 0.50,380 360 340

IIIB–V, IVB–V – – – – – C32/40, C28/35, C25/30,0.40, 0.45, 0.50,380 360 340

4. Corrosion induced by chlorides from sea waterXS1 CEM I, IIA, IIB–S, SRPC – – – C45/50, C35/45, C32/40,

0.35, 0.45, 0.50,380 360 340

IIB–V, IIIA – – – C40/50, C32/40, C28/35,0.35, 0.45, 0.50,380 360 340

IIIB–V, IVB–V – – – C32/40, C28/35, C25/30,0.40, 0.50, 0.50,380 340 340

XS2 CEM I, IIA, IIB–S, SRPC – – – C40/50, C32/40, C28/35,0.40, 0.50, 0.55,380 340 320

IIB–V, IIIA – – – C35/45, C28/35, C25/30,0.40, 0.50, 0.55,380 340 320

IIIB–V, IVB–V – – – C32/40, C25/30, C20/25,0.40, 0.50, 0.55,380 340 320

XS3 CEM I, IIA, IIB–S, SRPC – – – – – – C45/55, C40/50,0.35, 0.40,380 380

IIB–V, IIIA – – – – – C35/45, C32/40, C28/35,0.40, 0.45, 0.50,380 360 340

IIIB–V, IVB – – – – – C32/40, C28/35, C25/30,0.40, 0.45, 0.50,380 360 340

1Cement or combination types:

CEM I Portland cementIIA Portland cement with 6–20% pfa, ggbs or 20% limestoneIIB Portland cement with 21–35% pfa or ggbsIIIA Portland cement with 36–65% ggbsIIIB Portland cement with 66–80% pfa or ggbsIVB Portland cement with 36–55% pfaSRPC Sulphate resisting Portland cement–S ground granulated blast furnace slag (ggbs)–V pulverised fly ash (pfa)

9780415467193_C03a 9/3/09, 12:41 PM40

41


conditions of very low humidity, assumed to beless than about 35 per cent, seldom exist in prac-tice. The table also distinguishes between chloridesderived from sea-water and chlorides derived fromother sources, presumably rock salt, which is usedas a de-icing agent during winter maintenance. Asnoted above, these mechanisms may occur singlyor in combination. For example, an external ele-ment of a building structure may be susceptible tocarbonation and freeze thaw attack, i.e. exposureclasses XC4 + XF1. Similarly, coastal structuresmay be vulnerable to both chloride attack andfreezing, i.e. exposure classes XS1 + XF2. Clearly,the durability requirements should be based on themost onerous condition.

Once the relevant environmental condition(s)have been identified, a minimum strength class andnominal depth of concrete cover to the reinforce-ment can be selected. Table 3.6 gives the nominal(i.e. minimum plus an allowance for deviation,normally assumed to be 10 mm) depths of concretecover to all reinforcement for specified cement/combination types and strength classes (for bothdesigned and designated concretes) required forexposure classes XC1-XC4, XD1-XD3 and XS1-XS3, for structures with an intended working lifeof at least 50 years. Reference should be made toTable A5 of BS 8500-1 for concrete covers for struc-tures with an intended working life of in excess of100 years.

It can be seen from Table 3.6 that for agiven level of protection, the permitted minimumquality of concrete decreases as the recommendednominal depth of concrete cover increases. More-over, for chloride-induced corrosion the permittedminimum strength class of concrete reduces withincreasing percentage of pulverised fuel ash (pfa)or ground granulated blastfurnace slag (ggbs). Forexample, for exposure class XD2 the minimumconcrete strength class is C40/50 if the pfa contentlies between 6–20 per cent (i.e. cement type IIA).However, a lower concrete strength class (C35/45)

is permitted if the pfa content is between 21–35per cent (i.e. cement type IIB) and lower still(C32/40) if the pfa content is between 35–55 percent (i.e. cement type IVB). Where concrete isvulnerable to freeze/thaw attack, i.e. exposureclasses XF1–XF4, the strength class of the con-crete must not generally fall below the values shownin Table 3.7.

Concrete in the ground (e.g. foundations) maybe subject to chemical attack, possibly due to thepresence of sulphates, magnesium or acids inthe soil and/or groundwater. Table 3.8 shows thenominal covers and design chemical (DC) anddesignated concrete classes (FND) for specifiedsoil chemical environments. The design procedureinvolves determining the class of aggressive chem-ical environment for concrete (ACEC) via limitson the sulphate and magnesium ion concentrationsand soil acidity. This is used together with theintended working life of the structure to determinethe DC class. Where the strength class of the con-crete exceeds C25/30, a designed concrete will haveto be specified and the concrete producer shouldbe advised of the DC class required. Otherwise,a designated (FND) concrete, with a minimumstrength of C25/30, can be specified. Where con-crete is cast directly against the earth the nominaldepth of concrete cover should be at least 75 mmwhereas for concrete cast against blinding it shouldbe at least 50 mm.

BS 8110 further recommends that the maximumsurface crack width should not exceed 0.3 mmin order to avoid corrosion of the reinforcing bars.This requirement will generally be satisfied byobserving the detailing rules given in BS 8110 withregard to:

1. minimum reinforcement areas;2. maximum clear spacing between reinforcing bars.

These requirements will be discussed individu-ally for beams, slabs and columns in sections 3.9.1.6,3.10.2.4 and 3.13.6, respectively.

Table 3.7 Minimum strength classes of concrete with 20 mm aggregatesto resist freeze thaw attack (based on Table A8 BS 8500-1)

Exposure class: XF1 XF2 XF3 XF4

Indicative Strength Classes:3.5% air-entrainment C28/35 C32/40 C40/50 C40/50No air-entrainment RC25/30 RC25/30 RC25/30 RC28/35

9780415467193_C03a 9/3/09, 12:41 PM41


42

Tab

le 3

.8A

CE

C c

lass

es a

nd a

ssoc

iate

d no

min

al c

over

s an

d D

C o

r de

sign

ated

con

cret

es f

or s

truc

ture

s w

ith

an i

nten

ded

wor

king

life

of a

t le

ast

50 y

ears

(ba

sed

on T

able

s A

.2,

A.9

and

A.1

0 of

BS

850

0-1

)

Sul

phat

e an

d m

agne

sium

Des

ign

Nat

ural

soi

lB

row

n fiel

dA

CE

C-

Low

est

DC

/FN

D

2:1

wat

er/s

oil

Gro

undw

ater

Tot

alsu

lpha

teS

tatic

Mob

ileS

tatic

Mob

ilecl

ass

nom

inal

clas

sa

Ext

ract

pote

ntia

lcl

ass

wat

erw

ater

wat

erw

ater

cove

rs

sulp

hate

(mm

)

SO

4M

gS

O4

Mg

SO

4

mg/

lm

g/l

mg/

lm

g/l

%

1600

to

–15

00 t

o–

0.7–

1.2

DS

-3>3

.5–

>5.5

–A

C-2

s50

b , 75

cD

C-2

(F

ND

2)30

0030

00–

>5.5

–>6

.5A

C-3

50b ,

75c

DC

-3 (

FN

D3)

2.5

–3.5

–2.

5–5

.5–

AC

-3s

50b ,

75c

DC

-3 (

FN

D3)

–2.

5–5

.5–

5.6

–6.

5A

C-4

50b ,

75c

DC

-4 (

FN

D4)

––

–2.

5–5

.5A

C-5

50b ,

75c

DC

-4 (

FN

D4)

+ A

PM

3d

3100

to

≤120

031

00 t

o≤1

000

1.3

–2.4

DS

-4>3

.5–

>5.5

–A

C-3

s50

b , 75

cD

C-3

(FN

D3)

6000

6000

–>5

.5–

>6.5

AC

-450

b , 75

cD

C-4

(F

ND

4)2.

5–3

.5–

2.5

–5.5

–A

C-4

s50

b , 75

cD

C-4

(F

ND

4)–

2.5

–5.5

–2.

5–

6.5

AC

-550

b , 75

cD

C-4

(F

ND

4)+

AP

M3d

a For

str

uctu

res

wit

h at

lea

st 5

0 ye

ars’

wor

king

lif

eb F

or c

oncr

ete

cast

aga

inst

blin

ding

c For

con

cret

e ca

st d

irec

tly

agai

nst

the

soil

d Add

itio

nal

Pro

tect

ion

Mea

sure

(A

PM

) 3

– pr

ovid

e su

rfac

e pr

otec

tion

9780415467193_C03a 9/3/09, 12:41 PM42

43

Example 3.1 Selection of minimum strength class and nominal concretecover to reinforcement (BS 8110)

Assuming a design life of 50 years, determine the minimum concrete strength classes of concrete and the associatednominal covers to reinforcement at locations 1–4 for the structure shown in Fig. 3.10. List any assumptions.


4

3

2

1

Basementcar park

LOCATION 1Assume concrete column is exposed to rain and freezing.

Therefore, design the column for exposure class XC4 and XF1 (Table 3.5).From Table 3.7 the minimum strength class of concrete for class XF1 exposure is C28/35 and from Table 3.6 the

associated nominal cover to reinforcement for class XC4 exposure, cnom, is

cnom = 30 + ∆c = 30 + 10 = 40 mm

LOCATION 2Assume concrete beam is exposed to normal humidity.

Therefore, design the beam for exposure class XC1 (Table 3.5).From Table 3.6 the minimum strength class of concrete for class XC1 exposure is C20/25 and the associated

nominal cover to reinforcement, cnom, is

cnom = 15 + ∆c = 15 + 10 = 25 mm

LOCATION 3Clearly, the car-park slab is vulnerable to chloride attack but exposure class XD3 would seem to be too severe for a base-ment car park whereas exposure class XD1 is perhaps rather mild. As a compromise it is suggested that the minimumstrength class of concrete should be taken as C32/40 and the nominal cover to reinforcement, cnom, should be taken as

cnom = 30 + ∆c = 30 + 10 = 40 mm

LOCATION 4Assume non-aggressive soil conditions and that the concrete is cast directly against the soil.

Hence, design foundation for exposure class XC2 (Table 3.5).From Table 3.6 the minimum strength class of concrete for class XC2 exposure is C25/30 and the associated

nominal cover to reinforcement, cnom, is

cnom = 25 + ∆c = 25 + 10 = 35 mm ≥ 75 mm (since the concrete is cast directly against the ground).

Therefore cnom = 75 mm.

Fig. 3.10

9780415467193_C03a 9/3/09, 12:42 PM43


44

Table 3.9 Nominal cover to all reinforcement to meet specifiedperiods of fire resistance (based on Table 3.4, BS 8110)

Fire Nominal cover (mm)resistance

Beams Floors Columns(hours)

Simply Continuous Simply Continuoussupported supported

0.5 20 20 20 20 201.0 20 20 20 20 201.5 20 20 25 20 202.0 40 30 35 25 253.0 60 40 45 35 254.0 70 50 55 45 25

Fire Minimum dimension (mm)resistance(hours) Beam Floor Exposed

width thickness column width(b) (h) (b)

0.5 200 75 1501.0 200 95 2001.5 200 110 2502.0 200 125 3003.0 240 150 4004.0 280 170 450

Fig. 3.11 Minimum dimensions of reinforced concretemembers for fire resistance (based on Fig. 3.2, BS 8110).

Beams

b

Floors

Plane soffit h

Columns

b b

Fully exposed

3.8.2 FIRE PROTECTIONFire protection of reinforced concrete members islargely achieved by specifying limits for:

1. nominal thickness of cover to the reinforcement;2. minimum dimensions of members.

Table 3.9 gives the actual values of the nominaldepths of concrete covers to all reinforcement forspecified periods of fire resistance and membertypes. The covers in the table may need to beincreased because of durability considerations. Theminimum dimensions of members for fire resistanceare shown in Fig. 3.11.

Having discussed these more general aspectsrelating to structural design, the detailed design ofbeams is considered in the following section.

3.9 BeamsBeams in reinforced concrete structures can bedefined according to:

1. cross-section2. position of reinforcement3. support conditions.

Some common beam sections are shown inFig. 3.12. Beams reinforced with tension steel onlyare referred to as singly reinforced. Beams reinforcedwith tension and compression steel are termeddoubly reinforced. Inclusion of compression steelwill increase the moment capacity of the beam andhence allow more slender sections to be used. Thus,doubly reinforced beams are used in preference to

(a) (b)

Compressionsteel

Tension steel

Neutral

(c) (d)

axis

Fig. 3.12 Beam sections: (a) singly reinforced; (b) doublyreinforced; (c) T-section; (d) L-section.

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45

3.9.1 SINGLY REINFORCED BEAM DESIGNAll beams may fail due to excessive bending or shear.In addition, excessive deflection of beams must beavoided otherwise the efficiency or appearance ofthe structure may become impaired. As discussedin section 3.4, bending and shear are ultimate stateswhile deflection is a serviceabilty state. Generally,structural design of concrete beams primarily in-volves consideration of the following aspects whichare discussed next:

1. bending2. shear3. deflection.

3.9.1.1 Bending (clause 3.4.4.4, BS 8110)Consider the case of a simply supported, singlyreinforced, rectangular beam subject to a uniformlydistributed load ω as shown in Figs 3.15 and 3.16.

Fig. 3.13 Support conditions: (a) simply supported;(b) continuous.

(a) (b)

singly reinforced beams when there is some restric-tion on the construction depth of the section.

Under certain conditions, T and L beams aremore economical than rectangular beams sincesome of the concrete below the dotted line (neu-tral axis), which serves only to contain the tensionsteel, is removed resulting in a reduced unit weightof beam. Furthermore, beams may be simply sup-ported at their ends or continuous, as illustrated inFig. 3.13.

Figure 3.14 illustrates some of the notationused in beam design. Here b is the width of thebeam, h the overall depth of section, d the effectivedepth of tension reinforcement, d ′ the depth ofcompression reinforcement, As the area of tensionreinforcement and As′ the area of compressionreinforcement.

The following sub-sections consider the designof:

1. singly reinforced beams2. doubly reinforced beams3. continuous, L and T beams.

Fig. 3.16 Stress and strain distributions at section A-A: (a) section; (b) strains; (c) triangular (low strain);(d) rectangular parabolic (large strain); (e) equivalent rectangular.

ε cc

x

d

A sε st 1 4 4 4 4 2 4 4 4 4 3

Stress blocks

Neutral axis

0.67f cu

γm0.67f cu

γm

0.9x

(a) (b) (c) (d) (e)

b

d ′d

As′

As

h

Fig. 3.14 Notation.

Beams

A

A

Moment

Deflection

Compression zone

Tension zone

Cracks Neutral axis

Moment

ω

Fig. 3.15

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46

The load causes the beam to deflect downwards,putting the top portion of the beam into compres-sion and the bottom portion into tension. At somedistance x below the compression face, the sectionis neither in compression nor tension and thereforethe strain at this level is zero. This axis is normallyreferred to as the neutral axis.

Assuming that plane sections remain plane, thestrain distribution will be triangular (Fig. 3.16b).The stress distribution in the concrete above theneutral axis is initially triangular (Fig. 3.16c), forlow values of strain, because stress and strain aredirectly proportional (Fig. 3.7). The stress in theconcrete below the neutral axis is zero, however,since it is assumed that the concrete is cracked,being unable to resist any tensile stress. All thetensile stresses in the member are assumed to beresisted by the steel reinforcement and this isreflected in a peak in the tensile stress at the levelof the reinforcement.

As the intensity of loading on the beam increases,the mid-span moment increases and the distributionof stresses changes from that shown in Fig. 3.16cto 3.16d. The stress in the reinforcement increaseslinearly with strain up to the yield point. Thereafterit remains at a constant value (Fig. 3.9). However,as the strain in the concrete increases, the stressdistribution is assumed to follow the parabolic formof the stress–strain relationship for concrete undercompression (Fig. 3.7).

The actual stress distribution at a given sectionand the mode of failure of the beam will dependupon whether the section is (1) under-reinforcedor (2) over-reinforced. If the section is over-reinforced the steel does not yield and the failuremechanism will be crushing of the concrete due toits compressive capacity being exceeded. Steel isexpensive and, therefore, over-reinforcing will leadto uneconomical design. Furthermore, with this typeof failure there may be no external warning signs;just sudden, catastrophic collapse.

If the section is under-reinforced, the steel yieldsand failure will again occur due to crushing ofthe concrete. However, the beam will show con-siderable deflection which will be accompaniedby severe cracking and spalling from the tensionface thus providing ample warning signs of failure.Moreover, this form of design is more economicalsince a greater proportion of the steel strength isutilised. Therefore, it is normal practice to designsections which are under-reinforced rather thanover-reinforced.

In an under-reinforced section, since the rein-forcement will have yielded, the tensile force in the

steel (Fst) at the ultimate limit state can be readilycalculated using the following:

Fst = design stress × area

=

f Ay s

msγ (using equation 3.4) (3.5)

where

fy = yield stressA s = area of reinforcementγms = factor of safety for reinforcement (= 1.15)

However, it is not an easy matter to calculatethe compressive force in the concrete because ofthe complicated pattern of stresses in the concrete.To simplify the situation, BS 8110 replaces therectangular–parabolic stress distribution with anequivalent rectangular stress distribution (Fig. 3.16e).And it is the rectangular stress distribution whichis used in order to develop the design formulaefor rectangular beams given in clause 3.4.4.4 ofBS 8110. Specifically, the code gives formulae forthe following design parameters which are derivedbelow:

1. ultimate moment of resistance2. area of tension reinforcement3. lever arm.

(i) Ultimate moment of resistance, Mu. Con-sider the singly reinforced beam shown in Fig. 3.17.The loading on the beam gives rise to an ultimatedesign moment (M ) at mid-span. The resultingcurvature of the beam produces a compression forcein the concrete (Fcc) and a tensile force in the rein-forcement (Fst). Since there is no resultant axialforce on the beam, the force in the concrete mustequal the force in the reinforcement:

Fcc = Fst (3.6)

These two forces are separated by a distance z,the moment of which forms a couple (Mu) whichopposes the design moment. For structural stabil-ity Mu ≥ M where

Mu = Fccz = Fstz (3.7)

From the stress block shown in Fig. 3.17(c)

Fcc = stress × area

=

0 670 9

..

fxbcu

mcγ(3.8)

and

z = d − 0.9x/2 (3.9)

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47

Fig. 3.18 Lever-arm curve.

Rat

io z

/d

0.95

0.9

0.85

0.8

0.7740 0.042 0.05 0.1 0.15 0.156

M /bd 2f cu

Beams

Fig. 3.17 Ultimate moment of resistance for singly reinforced section.

Fst

(c)(b)(a)

d

x

Neutral axis

0.9xFcc

z

0.67fcu

γmb

In order to ensure that the section is under-reinforced, BS 8110 limits the depth of the neutralaxis (x) to a maximum of 0.5d, where d is theeffective depth (Fig. 3.17(b)). Hence

x ≤ 0.5d (3.10)

By combining equations 3.7–3.10 and puttingγmc = 1.5 (Table 3.3) it can be shown that the ulti-mate moment of resistance is given by:

Mu = 0.156fcubd2 (3.11)

Note that Mu depends only on the properties ofthe concrete and not the steel reinforcement. Pro-vided that the design moment does not exceed Mu

(i.e. M ≤ Mu), a beam whose section is singly rein-forced will be sufficient to resist the design moment.The following section derives the equation neces-sary to calculate the area of reinforcement neededfor such a case.

(ii) Area of tension reinforcement, As. At thelimiting condition Mu = M, equation 3.7 becomes

M = Fst·z

=

f Azy s

msγ(from equation 3.5)

Rearranging and putting γms = 1.15 (Table 3.3)gives

As =

Mf z0 87. y

(3.12)

Solution of this equation requires an expression forz which can either be obtained graphically (Fig. 3.18)or by calculation as discussed below.

(iii) Lever arm, z. At the limiting conditionMu = M, equation 3.7 becomes

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48

M = Fccz =

0 67. fcu

mcγ0.9bxz (from equation 3.8)

= 0.4fcubzx (putting γmc = 1.5)

= 0.4fcubz2

( ).

d z−0 9

(from equation 3.9)

=

89

fcubz(d − z)

Dividing both sides by fcubd 2 gives

Mf bdcu

2

89

= (z/d )(1 − z/d )

Substituting K =

Mf bdcu

2 and putting zo = z/d gives

0 = zo2 − zo + 9K/8

This is a quadratic and can be solved to give

zo = z/d = 0.5 + ( . / . )0 25 0 9− K

This equation is used to draw the lever arm curveshown in Fig. 3.18, and is usually expressed in thefollowing form

z = d[0.5 + ( . / . )0 25 0 9− K ] (3.13)

Once z has been determined, the area of tensionreinforcement, As, can be calculated using equation3.12. In clause 3.4.4.1 of BS 8110 it is noted thatz should not exceed 0.95d in order to give a reason-able concrete area in compression. Moreover itshould be remembered that equation 3.12 can onlybe used to determine As provided that M ≤ Mu orK ≤ K ′ where

K =

Mf bdcu

2and K ′ =

Mf bd

u

cu2

To summarise, design for bending requiresthe calculation of the maximum design moment(M) and corresponding ultimate moment of re-sistance of the section (Mu). Provided M ≤ Mu orK ≤ K ′, only tension reinforcement is needed andthe area of steel can be calculated using equation3.12 via equation 3.13. Where M > Mu the de-signer has the option to either increase the sectionsizes (i.e. M ≤ Mu) or design as a doubly reinforcedsection. The latter option is discussed more fullyin section 3.9.2.

Example 3.2 Design of bending reinforcement for a singly reinforcedbeam (BS 8110)

A simply supported rectangular beam of 7 m span carries characteristic dead (including self-weight of beam), gk, andimposed, qk, loads of 12 kNm−1 and 8 kNm−1 respectively (Fig. 3.19). The beam dimensions are breadth, b, 275 mm andeffective depth, d, 450 mm. Assuming the following material strengths, calculate the area of reinforcement required.

fcu = 30 Nmm−2

fy = 500 Nmm−2

Ultimate load (w) = 1.4gk + 1.6qk

= 1.4 × 12 + 1.6 × 8 = 29.6 kNm−1

Design moment (M ) =

wl2 2

829 6 7

8

. =

×= 181.3 kNm

Ultimate moment of resistance (Mu) = 0.156fcubd2

= 0.156 × 30 × 275 × 4502 × 10−6 = 260.6 kNm

Fig. 3.19

b = 275

d = 450qk = 8 kN m−1

gk = 12 kN m−1

7 m

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49

3.9.1.2 Design chartsAn alternative method of determining the area oftensile steel required in singly reinforced rectangu-lar beams is by using the design charts given inPart 3 of BS 8110. These charts are based on therectangular–parabolic stress distribution for con-crete shown in Fig. 3.16(d) rather than the simpli-fied rectangular distribution and should thereforeprovide a more economical estimate of the requiredarea of steel reinforcement. However, BSI issuedthese charts when the preferred grade of reinforce-ment was 460, not 500, and use of these chartswill therefore in fact overestimate the required ten-sile steel area by around 10 per cent.

A modified version of chart 2 which is appropri-ate for use with grade 500 reinforcement is shownFig. 3.20.

Table 3.10 Cross-sectional areas of groups of bars (mm2)

Bar size Number of bars (mm)

1 2 3 4 5 6 7 8 9 10

6 28.3 56.6 84.9 113 142 170 198 226 255 2838 50.3 101 151 201 252 302 352 402 453 503

10 78.5 157 236 314 393 471 550 628 707 785

12 113 226 339 452 566 679 792 905 1020 113016 201 402 603 804 1010 1210 1410 1610 1810 201020 314 628 943 1260 1570 1890 2200 2510 2830 3140

25 491 982 1470 1960 2450 2950 3440 3930 4420 491032 804 1610 2410 3220 4020 4830 5630 6430 7240 804040 1260 2510 3770 5030 6280 7540 8800 10100 11300 12600

The design procedure involves the followingsteps:

1. Check M ≤ Mu.2. Select appropriate chart from Part 3 of BS 8110

based on the grade of tensile reinforcement.3. Calculate M/bd2.4. Plot M/bd 2 ratio on chart and read off corres-

ponding 100As/bd value using curve appropriateto grade of concrete selected for design.

5. Calculate As.

Using the figures given in Example 3.2,

M = 181.3 kNm ≤ Mu = 260.6 kNm

Mbd2

6

2

181 3 10275 450

3 26 .

.=×

×=

Beams

Example 3.2 continuedSince Mu > M design as a singly reinforced beam.

K =

Mf bdcu

2

6

2

181 3 1030 275 450

.

=

×× ×

= 0.1085

z = d[0.5 + ( . / . )0 25 0 9− K

= 450[0.5 + ( . . / . )0 25 0 1085 0 9− ]

= 386.8 mm ≤ 0.95d (= 427.5 mm) OK.

A s =

Mf z0 87

181 3 100 87 500 386 8

6

.

. . .y

=×

× ×= 1078 mm2

For detailing purposes this area of steel has to be transposed into a certain number of bars of a given diameter.This is usually achieved using steel area tables similar to that shown in Table 3.10. Thus it can be seen that four20 mm diameter bars have a total cross-sectional area of 1260 mm2 and would therefore be suitable. Hence provide4H20. (N.B. H refers to high yield steel bars (fy = 500 Nmm−2); R refers to mild steel bars (fy = 250 Nmm−2).

9780415467193_C03a 9/3/09, 12:43 PM49


50

As

fy 500

b

xd

M/b

d2

(N m

m−2

)

7

6

5

4

3

2

1

40

30

35

25

f cu

(N m

m−2

)

0 0.5 1.0 1.5 2.0

100As/bd

From Fig. 3.20, using the fcu = 30 N/mm2 curve

1000 87

Abd

s .=

Hence, As = 1076 mm2

Therefore provide 4H20 (A s = 1260 mm2)

3.9.1.3 Shear (clause 3.4.5, BS 8110)Another way in which failure of a beam may ariseis due to its shear capacity being exceeded. Shearfailure may arise in several ways, but the two prin-cipal failure mechanisms are shown in Fig. 3.21.

With reference to Fig. 3.21(a), as the loadingincreases, an inclined crack rapidly developsbetween the edge of the support and the load point,resulting in splitting of the beam into two pieces.This is normally termed diagonal tension failure andcan be prevented by providing shear reinforcement.

The second failure mode, termed diagonal com-pression failure (Fig. 3.21(b)), occurs under theaction of large shear forces acting near the support,resulting in crushing of the concrete. This typeof failure is avoided by limiting the maximumshear stress to 5 N/mm2 or 0.8 fcu , whichever isthe lesser.

The design shear stress, υ, at any cross-sectioncan be calculated from:

υ = V/bd (3.14)

whereV design shear force due to ultimate loadsb breadth of sectiond effective depth of section

In order to determine whether shear reinforcementis required, it is necessary to calculate the shear re-sistance, or using BS 8110 terminology the designconcrete shear stress, at critical sections along thebeam. The design concrete shear stress, υc, is foundto be composed of three major components, namely:

1. concrete in the compression zone;2. aggregate interlock across the crack zone;3. dowel action of the tension reinforcement.

The design concrete shear stress can be deter-mined using Table 3.11. The values are in termsof the percentage area of longitudinal tension rein-forcement (100As/bd ) and effective depth of the

Fig. 3.20 Design chart for singly reinforced beam (based on chart No. 2, BS 8110: Part 3).

(a) (b)

Fig. 3.21 Types of shear failure: (a) diagonal tension;(b) diagonal compression.

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51

Table 3.11 Values of design concrete shear stress, υc (N/mm2) forfcu = 25 N/mm2 concrete (Table 3.8, BS 8110)

Effective depth (d) mm

125 150 175 200 225 250 300 ≥ 400

≤ 0.15 0.45 0.43 0.41 0.40 0.39 0.38 0.36 0.340.25 0.53 0.51 0.49 0.47 0.46 0.45 0.43 0.400.50 0.57 0.64 0.62 0.60 0.58 0.56 0.54 0.500.75 0.77 0.73 0.71 0.68 0.66 0.65 0.62 0.571.00 0.84 0.81 0.78 0.75 0.73 0.71 0.68 0.631.50 0.97 0.92 0.89 0.86 0.83 0.81 0.78 0.722.00 1.06 1.02 0.98 0.95 0.92 0.89 0.86 0.80≥ 3.00 1.22 1.16 1.12 1.08 1.05 1.02 0.98 0.91

The former is the most widely used method andwill therefore be the only one discussed here. Thefollowing section derives the design equations forcalculating the area and spacing of links.

(i) Shear resistance of links. Consider a rein-forced concrete beam with links uniformly spacedat a distance sv, under the action of a shear force V.The resulting failure plane is assumed to be in-clined approximately 45° to the horizontal as shownin Fig. 3.22.

The number of links intersecting the potentialcrack is equal to d/sv and it follows therefore thatthe shear resistance of these links, Vlink, is given by

Vlink = number of links × total cross-sectionalarea of links (Fig. 3.23) × design stress

= (d/sv) × Asv × 0.87fyv

Fig. 3.23 A sv for varying shear reinforcement arrangements.

Asv = 2AC

πΦ2

4DF Asv = 4A

CπΦ2

4DF

Φ − diameter of links

section (d ). The table assumes that cube strengthof concrete is 25 Nmm−2. For other values of cubestrength up to a maximum of 40 Nmm−2, the de-sign shear stresses can be determined by multiply-ing the values in the table by the factor ( fcu/25)1/3.

Generally, where the design shear stress exceedsthe design concrete shear stress, shear reinforce-ment will be needed. This is normally done byproviding

1. vertical shear reinforcement commonly referredto as ‘links’

2. a combination of vertical and inclined (or bent-up) bars as shown below.

Beams

sv

d

d45°

Fig. 3.22 Shear resistance of links.

100Abd

s

Inclined shearreinforcement

Vertical shearreinforcement

45°

Beam with vertical and inclined shear reinforcement.

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52

gk = 10 kN m−1

qk varies

225 mm225 mm

6 m

b = 325

d = 547fcu = 25 N mm−2

4H25 (As = 1960 mm2)

The shear resistance of concrete, Vconc, can becalculated from

Vconc = υcbd (using equation 3.14)

The design shear force due to ultimate loads, V,must be less than the sum of the shear resistanceof the concrete (Vconc) plus the shear resistance ofthe links (Vlink), otherwise failure of the beam mayarise. Hence

V ≤ Vconc + Vlink

≤ υcbd + (d/sv)Asv0.87fyv

Dividing both sides by bd gives

V/bd ≤ υc + (1/bsv)Asv0.87fyv

From equation 3.14

υ ≤ υc + (1/bsv)Asv0.87fyv

rearranging gives

As

bf

sv

v

c

yv

( ).

=−ν ν

0 87(3.15)

Where (υ − υc) is less than 0.4 N/mm2 then linksshould be provided according to

As

bf

sv

v yv

.

.=

0 40 87

(3.16)

Equations 3.15 and 3.16 provide a basis for cal-culating the minimum area and spacing of links.The details are discussed next.

(ii) Form, area and spacing of links. Shearreinforcement should be provided in beams ac-cording to the criteria given in Table 3.12.

Thus where the design shear stress is less thanhalf the design concrete shear stress (i.e. υ < 0.5υc),no shear reinforcement will be necessary although,in practice, it is normal to provide nominal linksin all beams of structural importance. Where0.5υc < υ < (υc + 0.4) nominal links based on equa-tion 3.16 should be provided. Where υ > υc + 0.4,design links based on equation 3.15 should beprovided.

BS 8110 further recommends that the spacing oflinks in the direction of the span should not exceed0.75d. This will ensure that at least one link crossesthe potential crack.

Example 3.3 Design of shear reinforcement for a beam (BS 8110)Design the shear reinforcement for the beam shown in Fig. 3.24 using high yield steel (fy = 500 Nmm−2) links for thefollowing load cases:

(i) qk = 0(ii) qk = 10 kNm−1

(iii) qk = 29 kNm−1

(iv) qk = 45 kNm−1

Fig. 3.24

Table 3.12 Form and area of links in beams (Table 3.7, BS 8110)

Values of υ (N/mm 2) Area of shear reinforcement to be provided

υ < 0.5υc throughout the beam No links required but normal practice to provide nominal links inmembers of structural importance

0.5υc < υ < (υc + 0.4) Nominal (or minimum) links for whole length of beam A

bsfsv

v

yv

0.40.87

≥

(υc + 0.4) < υ < 0 8. f cu or 5 N/mm2 Design links A

bsfsv

v c

yv0

( ).

≥−υ υ

87

9780415467193_C03a 9/3/09, 12:44 PM52

53

Beams

Example 3.3 continuedDESIGN CONCRETE SHEAR STRESS, υc

100 100 1960325 547

1 10A

bds

.=××

=

From Table 3.11, υc = 0.65 Nmm−2 (see below)

(I) qk = 0Design shear stress (υυυυυ)

100A s Effective depth (mm) bd

300 ≥ 400

Nmm−2 Nmm−2

1.00 0.68 0.631.10 0.651.50 0.78 0.72

Total ultimate load, W, is

W = (1.4gk + 1.6qk)span = (1.4 × 10 + 1.6 × 0)6 = 84 kN

Since beam is symmetrically loaded

RA = RB = W/2 = 42 kN

Ultimate shear force (V ) = 42 kN and design shear stress, υ, is

υ

= =××

Vbd

42 10325 547

3

= 0.24 Nmm−2

Diameter and spacing of linksBy inspection

υ < υc/2

i.e. 0.24 Nmm−2 < 0.32 Nmm−2. Hence from Table 3.12, shear reinforcement may not be necessary.

gk = 10 kN m−1

RA RB

V

V

6 m

Shear force diagram

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54

Table 3.13 Values of A sv/s v

Diameter Spacing of links (mm)(mm)

85 90 100 125 150 175 200 225 250 275 300

8 1.183 1.118 1.006 0.805 0.671 0.575 0.503 0.447 0.402 0.366 0.33510 1.847 1.744 1.57 1.256 1.047 0.897 0.785 0.698 0.628 0.571 0.52312 2.659 2.511 2.26 1.808 1.507 1.291 1.13 1.004 0.904 0.822 0.75316 4.729 4.467 4.02 3.216 2.68 2.297 2.01 1.787 1.608 1.462 1.34


W = (1.4gk + 1.6qk)span = (1.4 × 10 + 1.6 × 10)6 = 180 kN


RA = RB = W/2 = 90 kN


υ

.= =××

=Vbd

90 10325 547

0 513

Nmm−2


υc /2 < υ < (υc + 0.4)

i.e. 0.32 < 0.51< 1.05. Hence from Table 3.12, provide nominal links for whole length of beam according to

As

bf

sv

v yv

.

.

. .

= =××

0 40 87

0 4 3250 87 500

This value has to be translated into a certain bar size and spacing of links and is usually achieved using shearreinforcement tables similar to Table 3.13. The spacing of links should not exceed 0.75d = 0.75 × 547 = 410 mm.From Table 3.13 it can be seen that 8 mm diameter links spaced at 300 mm centres provide a A sv/s v ratio of 0.335 andwould therefore be suitable. Hence provide H8 links at 300 mm centres for whole length of beam.

gk = 10 kN m−1 qk = 10 kN m−1

RA RB

V

V

6 m

Example 3.3 continued(II) qk = 10 kNm−1

Design shear stress (υυυυυ)

9780415467193_C03a 9/3/09, 12:45 PM54

55

21H8 links at 300

2H12 (hanger bars)

H8 links

4H25

gk = 10 kN m−1

RA

6 mRB

VV

qk = 45 kN m−1


W = (1.4gk + 1.6qk)span = (1.4 × 10 + 1.6 × 29)6 = 362.4 kN


RA = RB = W/2 = 181.2 kN

Ultimate shear force (V ) = 181.2 kN and design shear stress, υ, is

υ

.

= =×

×Vbd

181 2 10325 547

3

= 1.02 Nmm−2


υc /2 < υ < (υc + 0.4)

i.e. 0.32 < 1.02 < 1.05. Hence from Table 3.12, provide nominal links for whole length of beam according to

As

bf

sv

v yv

.

.

. .

= =××

0 40 87

0 4 3250 87 500

= 0.3

Therefore as in case (ii) (qk = 10 kNm−1), provide H8 links at 300 mm centres.

(IV) qk = 45 kNm−1

Beams


gk = 10 kN m−1qk = 29 kN m−1

RA RB

V

V

6 m

(III) qk = 29 kNm−1

Design shear stress (v)

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56

7H8 at 150 13H8 at 300

Numberof links

Grade of steelDiameterof links

7H8 at 150

Centre-to-centredistance betweenlinks 4H25

H8 links

2H12 (hanger bars)

Example 3.3 continuedDesign shear stress (v)Total ultimate load, W, is

W = (1.4gk + 1.6qk)span = (1.4 × 10 + 1.6 × 45)6 = 516 kN


RA = RB = W/2 = 258 kN


υ

.= =××

=Vbd

258 10325 547

1 453

Nmm−2 < permissible = 0 8 25. = 4 Nmm−2

Diameter and spacing of linksWhere υ < (υc + 0.4) = 0.65 + 0.4 = 1.05 Nmm−2, nominal links are required according to

As

bf

sv

v yv

.

.

. .

.= =××

=0 4

0 870 4 3250 87 500

0 3

Hence, from Table 3.13, provide H8 links at 300 mm centres where υ < 1.05 Nmm−2, i.e. 2.172 m either side of themid-span of beam.

x/1.05 = 3/1.45

v =

1.45

N m

m−2

v3 m 3 m

xx

1.05 N mm−2

x = 2.172 m

Where υ > (υ c + 0.4) = 1.05 Nmm−2 design links required according to

As

bf

csv

v yv

( )

.

( . . ).

.=−

=−

×=

υ υ0 87

3251 45 0 650 87 500

0 598

Hence, from Table 3.13, provide H8 links at 150 mm centres (A sv/s v = 0.671) where v > 1.05 Nmm−2, i.e. 0.828 m infrom both supports.

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57

Table 3.14 Basic span /effective depth ratiofor rectangular or flanged beams (Table 3.9,BS 8110)

Support conditions Rectangular Flanged beams withsections

width of beamwidth of flange

.≤ 0 3

Cantilever 7 5.6Simply supported 20 16.0Continuous 26 20.8

3.9.1.4 Deflection (clause 3.4.6, BS 8110)In addition to checking that failure of the memberdoes not arise due to the ultimate limit states ofbending and shear, the designer must ensure thatthe deflections under working loads do not adverselyaffect either the efficiency or appearance of thestructure. BS 8110 describes the following criteriafor ensuring the proper performance of rectangularbeams:

1. Final deflection should not exceed span/250.2. Deflection after construction of finishes and

partitions should not exceed span/500 or 20 mm,whichever is the lesser, for spans up to 10 m.

However, it is rather difficult to make accuratepredictions of the deflections that may arise in con-crete members principally because the member maybe cracked under working loads and the degreeof restraint at the supports is uncertain. Therefore,BS 8110 uses an approximate method based onpermissible ratios of the span/effective depth. Beforediscussing this method in detail it is worth clarifyingwhat is meant by the effective span of a beam.

(i) Effective span (clause 3.4.1.2, BS 8110). Allcalculations relating to beam design should be basedon the effective span of the beam. For a simplysupported beam this should be taken as the lesserof (1) the distance between centres of bearings, A,or (2) the clear distance between supports, D, plusthe effective depth, d, of the beam (Fig. 3.25). Fora continuous beam the effective span should nor-mally be taken as the distance between the centresof supports.

(ii) Span/effective depth ratio. Generally, thedeflection criteria in (1) and (2) above will be satis-fied provided that the span/effective depth ratio ofthe beam does not exceed the appropriate limitingvalues given in Table 3.14. The reader is referredto the Handbook to BS 8110 which outlines thebasis of this approach.

The span/effective depth ratio given in the tableapply to spans up to 10 m long. Where the spanexceeds 10 m, these ratios should be multiplied by

Table 3.15 Modification factors for compressionreinforcement (Table 3.11, BS 8110)

100 ′Abd

s , prov Factor

0.00 1.00.15 1.050.25 1.080.35 1.100.5 1.140.75 1.201 1.251.5 1.332.0 1.402.5 1.45≥ 3.0 1.5

d

D

A

Fig. 3.25 Effective span of simply supported beam.

10/span (except for cantilevers). The basic ratiosmay be further modified by factors taken fromTables 3.15 and 3.16, depending upon the amountof compression and tension reinforcement respect-ively. Deflection is usually critical in the design ofslabs rather than beams and, therefore, modifica-tions factors will be discussed more fully in thecontext of slab design (section 3.10).

3.9.1.5 Member sizingThe dual concepts of span/effective depth ratio andmaximum design concrete shear stress can be usednot only to assess the performance of memberswith respect to deflection and shear but also forpreliminary sizing of members. Table 3.17 givesmodified span/effective depth ratios for estimatingthe effective depth of a concrete beam providedthat its span is known. The width of the beamcan then be determined by limiting the max-imum design concrete shear stress to around (say)1.2 Nmm−2.

Beams

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58

Table 3.16 Modification factors for tension reinforcement (based on Table 3.10, BS 8110)

Service stress M/bd 2

0.50 0.75 1.00 1.50 2.00 3.00 4.00 5.00 6.00

100 2.00 2.00 2.00 1.86 1.63 1.36 1.19 1.08 1.01150 2.00 2.00 1.98 1.69 1.49 1.25 1.11 1.01 0.94

( fy = 250) 167 2.00 2.00 1.91 1.63 1.44 1.21 1.08 0.99 0.92200 2.00 1.95 1.76 1.51 1.35 1.14 1.02 0.94 0.88250 1.90 1.70 1.55 1.34 1.20 1.04 0.94 0.87 0.82300 1.60 1.44 1.33 1.16 1.06 0.93 0.85 0.80 0.76

( fy = 500) 323 1.41 1.28 1.18 1.05 0.96 0.86 0.79 0.75 0.72

Note 1. The values in the table derive from the equation:

Modification factor

= +−

+

≤ . ( )

.

.0 55477

120 0 9

2 0f

M

bd

s

2

(equation 7)

wherefs is the design service stress in the tension reinforcementM is the design ultimate moment at the centre of the span or, for a cantilever, at the support.Note 2. The design service stress in the tension reinforcement may be estimated from the equation:

f

f A

Asy s,req

s,prov b

= × ×5 1*

8 β(equation 8)

where β b is the percentage of moment redistribution, equal to 1 for simply supported beams.* As pointed out in Reynolds RC Designers Handbook the term 5/8 which is applicable to γms = 1.15 is given incorrectly as 2/3in BS 8110 which is applicable to γms = 1.05.

Table 3.17 Span/effective depth ratios forinitial design

Support condition Span/effective depth

Cantilever 6Simply supported 12Continuous 15

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59

Example 3.4 Sizing a concrete beam (BS 8110)A simply supported beam has an effective span of 8 m and supports characteristic dead (gk) and live (qk) loads of15 kNm−1 and 10 kNm−1 respectively. Determine suitable dimensions for the effective depth and width of the beam.

From Table 3.17, span/effective depth ratio for a simply supported beam is 12. Hence effective depth, d, is

d

span = = ≈

12800012

670 mm

Total ultimate load = (1.4gk + 1.6qk)span = (1.4 × 15 + 1.6 × 10)8 = 296 kNDesign shear force (V ) = 296/2 = 148 kN and design shear force, υ, is

υ

= =

×Vbd b

148 10670

3

Assuming υ is equal to 1.2 Nmm−2, this gives a beam width, b, of

b

Vd

.

= =××

=υ

148 10670 1 2

1853

mm

Hence a beam of width 185 mm and effective depth 670 mm would be suitable to support the given design loads.

qk = 10 kN m−1

gk = 15 kN m−1

8 m

Beams

3.9.1.6 Reinforcement details (clause 3.12,BS 8110)

The previous sections have covered much of thetheory required to design singly reinforced con-crete beams. However, there are a number of codeprovisions with regard to:

1. maximum and minimum reinforcement areas2. spacing of reinforcement3. curtailment and anchorage of reinforcement4. lapping of reinforcement.

These need to be taken into account since theymay affect the final design.

1. Reinforcement areas (clause 3.12.5.3 and3.12.6.1, BS 8110). As pointed out in section 3.8,there is a need to control cracking of the concretebecause of durability and aesthetics. This is usuallyachieved by providing minimum areas of reinforce-ment in the member. However, too large an areaof reinforcement should also be avoided since itwill hinder proper placing and adequate compactionof the concrete around the reinforcement.

For rectangular beams with overall dimensions band h, the area of tension reinforcement, As, shouldlie within the following limits:

0.24%bh ≤ As ≤ 4%bh when fy = 250 Nmm−2

0.13%bh ≤ As ≤ 4%bh when fy = 500 Nmm−2

2. Spacing of reinforcement (clause 3.12.11.1,BS 8110). BS 8110 specifies minimum and max-imum distances between tension reinforcement.The actual limits vary, depending upon the gradeof reinforcement. The minimum distance is basedon the need to achieve good compaction of the con-crete around the reinforcement. The limits on themaximum distance between bars arise from the needto ensure that the maximum crack width does notexceed 0.3 mm in order to prevent corrosion ofembedded bars (section 3.8).

For singly reinforced simply supported beamsthe clear horizontal distance between tension bars,sb, should lie within the following limits:

hagg + 5 mm or bar size ≤ sb ≤ 280 mmwhen fy = 250 Nmm−2

hagg + 5 mm or bar size ≤ sb ≤ 155 mmwhen fy = 500 Nmm−2

where hagg is the maximum size of the coarseaggregate.

9780415467193_C03a 9/3/09, 12:47 PM59


60

Cut-off 50% of bars

50% 100% 50%

0.08� 0.08�

30% 30%

0.15�0.1�

100%

�

60%

0.15�

�45ø

0.25�

(a)

(b)

�

100%

Fig. 3.27 Simplified rules for curtailment of bars inbeams: (a) simply supported ends; (b) continuous beam.

12Φ

CL

12Φd2

+

(b)(a)

Fig. 3.28 Anchorage requirements at simple supports.

x = 0.146�

As/2 As

�

ω

B

ω�2

16ω�2ω�2

81.6

A

As/2

Fig. 3.26

3. Curtailment and anchorage of bars (clause3.12.9, BS 8110). The design process for simplysupported beams, in particular the calculations re-lating to the design moment and area of bendingreinforcement, is concentrated at mid-span. How-ever, the bending moment decreases either side ofthe mid-span and it follows, therefore, that it shouldbe possible to reduce the corresponding area ofbending reinforcement by curtailing bars. For thebeam shown in Fig. 3.26, theoretically 50 per centof the main steel can be curtailed at points A andB. However, in order to develop the design stressin the reinforcement (i.e. 0.87fy at mid-span), thesebars must be anchored into the concrete. Except atend supports, this is normally achieved by extend-ing the bars beyond the point at which they aretheoretically no longer required, by a distance equalto the greater of (i) the effective depth of the mem-ber and (ii) 12 times the bar size.

Where a bar is stopped off in the tension zone,e.g. beam shown in Fig. 3.26, this distance shouldbe increased to the full anchorage bond length inaccordance with the values given in Table 3.18.However, simplified rules for the curtailment ofbars are given in clause 3.12.10.2 of BS 8110. Theseare shown diagrammatically in Fig. 3.27 for simplysupported and continuous beams.

The code also gives rules for the anchorage ofbars at supports. Thus, at a simply supported endeach tension bar will be properly anchored providedthe bar extends a length equal to one of the fol-lowing: (a) 12 times the bar size beyond the centreline of the support, or (b) 12 times the bar sizeplus d/2 from the face of the support (Fig. 3.28).

Sometimes it is not possible to use straight barsdue to limitations of space and, in this case, an-chorage must be provided by using hooks or bendsin the reinforcement. The anchorage values of hooksand bends are shown in Fig. 3.29. Where hooks orbends are provided, BS 8110 states that they should

not begin before the centre of the support for rule(a) or before d/2 from the face of the support forrule (b).

4. Laps in reinforcement (clause 3.12.8, BS8110). It is not possible nor, indeed, practicableto construct the reinforcement cage for an indi-vidual element or structure without joining someof the bars. This is normally achieved by lappingbars (Fig. 3.30). Bars which have been joined inthis way must act as a single length of bar. Thismeans that the lap length should be sufficientlylong in order that stresses in one bar can be trans-ferred to the other.

9780415467193_C03a 9/3/09, 12:48 PM60

61

For mild steel bars minimum r = 2ΦFor high yield bars minimum r = 3Φ or4Φ for sizes 25 mm and above

Φr

4Φ

rΦ4Φ

(a)

(b)

+

+

Fig. 3.29 Anchorage lengths for hooks and bends(a) anchorage length for 90° bend = 4r but not greaterthan 12φ; (b) anchorage length for hook = 8r but notgreater than 24φ.

Table 3.18 Anchorage lengths as multiples ofbar size (based on Table 3.27, BS 8110)

LA

fcu = 25 30 35 40or more

Plain (250)Tension 43 39 36 34Compression 34 32 29 27Deformed Type 1 (500)Tension 55 50 47 44Compression 44 40 38 35Deformed Type 2 (500)Tension 44 40 38 35Compression 35 32 30 28

Example 3.5 Design of a simply supported concrete beam (BS 8110)A reinforced concrete beam which is 300 mm wide and 600 mm deep is required to span 6.0 m between the centresof supporting piers 300 mm wide (Fig. 3.31). The beam carries dead and imposed loads of 25 kNm−1 and 19 kNm−1

respectively. Assuming fcu = 30 Nmm−2, fy = fyv = 500 Nmm−2 and the exposure class is XC1, design the beam.

300

6 mA

A

300

q k = 19 kN m−1

g k = 25 kN m−1

Section A–A

b = 300

h = 600

Fig. 3.31

Beams

Kicker

Lap

d

Starter bars

Fig. 3.30 Lap lengths.

The minimum lap length should not be lessthan 15 times the bar diameter or 300 mm. Fortension laps it should normally be equal to the

tension anchorage length, but will often need tobe increased as outlined in clause 3.12.8.13 ofBS 8110. The anchorage length (L) is calculatedusing

L = LA × Φ (3.17)

whereΦ is the diameter of the (smaller) barLA is obtained from Table 3.18 and depends

upon the stress type, grade of concrete andreinforcement type.

For compression laps the lap length shouldbe at least 1.25 times the compression anchoragelength.

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62

Example 3.5 continuedDESIGN MOMENT, M

Loading

Dead

Self weight of beam = 0.6 × 0.3 × 24 = 4.32 kNm−1

Total dead load (gk) = 25 + 4.32 = 29.32 kNm−1

ImposedTotal imposed load (qk) = 19 kNm−1

Ultimate loadTotal ultimate load (W ) = (1.4gk + 1.6qk)span

= (1.4 × 29.32 + 1.6 × 19)6

= 428.7 kN

Design moment

Maximum design moment (M ) =

Wb8

428 7 68

321 5 .

. =×

= kN m

ULTIMATE MOMENT OF RESISTANCE, MU

Effective depth, d

d

c

Φ′

Φ/2

Assume diameter of main bars (Φ) = 25 mmAssume diameter of links (Φ′) = 8 mmFrom Table 3.6, cover for exposure class XC1 = 15 + ∆c = 25 mm.

d = h − c − Φ′ − Φ/2

= 600 − 25 − 8 − 25/2 = 554 mm

Ultimate momentMu = 0.156fcubd2 = 0.156 × 30 × 300 × 5542

= 430.9 × 106 Nmm = 430.9 kNm > M

Since Mu > M no compression reinforcement is required.

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63

Example 3.5 continuedMAIN STEEL, A s

K

Mf bd

.

.= =

×× ×

=cu

2

321 5 1030 300 554

0 1166

2

z = d K[ . ( . / . ) ]0 5 0 25 0 9+ − = 554 [ . ( . . / . )]0 5 0 25 0 116 0 9+ − = 470 mm

A

M

f zs

y

2mm .

.

. = =

×× ×

=0 87

321 5 10

0 87 500 4701573

6

Hence from Table 3.10, provide 4H25 (A s = 1960 mm2).

SHEAR REINFORCEMENT

Beams

Ultimate design load, W = 428.7 kN

Shear stress, υυυυυSince beam is symmetrically loaded

RA = RB = W/2 = 214.4 kN


υ

.

= =×

×Vbd

214 4 10300 554

3

= 1.29 Nmm−2 < permissible = 0.8 30 = 4.38 Nmm−2

Design concrete shear stress, υυυυυc

100 sAbd

.=××

=100 1960300 554

1 18

From Table 3.11,

υc = (30/25)1/3 × 0.66 = 0.70 Nmm−2

Diameter and spacing of linksWhere υ < (υc + 0.4) = 0.7 + 0.4 = 1.1 Nmm−2, nominal links are required according to

A

sbf

s

y

n

n n

.

.

. .

.= =××

=0 4

0 870 4 3000 87 500

0 276

Hence from Table 3.13, provide H8 links at 300 mm centres where υ < 1.10 Nmm−2, i.e. 2.558 m either side of themid-span of beam.

v =

1.29

N m

m−2

1.10 N mm−2

x

3 m

x

3 m

v

x

1.10 1.29

32.558 m= =

RA

V

W

V

RB

9780415467193_C03a 9/3/09, 12:48 PM63


64

r = 100 mm

H25

100 mm

150 mmCL of support

H8 links

H12

300 mm

4H8 at 22514H8 at 3004H8 at 225

5700 mm

2H12

554 mm

46 mm

300 mm

4H25

H8 links

300 mm

Example 3.5 continuedWhere υ > (υc + 0.4) = 1.10 Nmm−2 design links required according to

As

bf

cs

y

(n

n n

)

.

( . . ).

.=−

=−

×=

υ υ0 87

300 1 29 0 700 87 500

0 407

Maximum spacing of links is 0.75d = 0.75 × 554 = 416 mm. Hence from Table 3.13, provide 8 mm diameter links at225 mm centres (A sv/s v = 0.447) where v > 1.10 Nmm−2, i.e. 0.442 m in from both supports.

EFFECTIVE SPANThe above calculations were based on an effective span of 6 m, but this needs to be confirmed. As stated in section3.9.1.4, the effective span is the lesser of (1) centre-to-centre distance between support, i.e. 6 m, and (2) cleardistance between supports plus the effective depth, i.e. 5700 + 554 = 6254 mm. Therefore assumed span length of6 m is correct.

DEFLECTIONActual span/effective depth ratio = 6000/554 = 10.8

M

bd2

.

.=

××

=321 5 10

300 5543 5

6

2

and from equation 8 (Table 3.16)

f f

A

As y

s,req

s,prov

2Nmm= × × = × × = − 5

8

5

8500

1573

1960251

From Table 3.14, basic span/effective depth ratio for a simply supported beam is 20 and from Table 3.16, modificationfactor ≈ 0.97. Hence permissible span/effective depth ratio = 20 × 0.97 = 19 > actual (= 10.8) and the beam thereforesatisfies the deflection criteria in BS 8110.

REINFORCEMENT DETAILSThe sketch below shows the main reinforcement requirements for the beam. For reasons of buildability, the actualreinforcement details may well be slightly different and the reader is referred to the following publications for furtherinformation on this point:

1. Designed and Detailed (BS 8110: 1997), Higgins, J.B. and Rogers, B.R., British Cement Association, 1989.2. Standard Method of Detailing Structural Concrete, the Concrete Society and the Institution of Structural Engin-

eers, London, 1989.

9780415467193_C03a 9/3/09, 12:49 PM64

65

4H25(A s = 1960 mm2)

30 mm cover

b = 300

h = 500

Beams

Example 3.6 Analysis of a singly reinforced concrete beam (BS 8110)A singly reinforced concrete beam in which fcu = 30 Nmm−2 and fy = 500 Nmm−2 contains 1960 mm2 of tensionreinforcement (Fig. 3.32). If the effective span is 7 m and the density of reinforced concrete is 24 kNm−3, calculate themaximum imposed load that the beam can carry assuming that the load is (a) uniformly distributed and (b) occurs asa point load at mid-span.

Fig. 3.32

(A) MAXIMUM UNIFORMLY DISTRIBUTED IMPOSED LOAD, qk

Moment capacity of section, M

Effective depth, d, is

d = h − cover − φ/2 = 500 − 30 − 25/2 = 457 mm

For equilibrium Fcc = Fst

0 67. fcu

mcg0.9xb = 0.87fy A s (assuming the steel has yielded)

0 67 301 5

. .×

0.9 × x × 300 = 0.87 × 500 × 1960 ⇒ x = 236 mm

z

Fcc

Fst

0.67fcu/γmcγ

0.9xx

d

b εcu = 0.0035ε

εstε

εyε

200 kNmm–2

fv/γms = 500/1.15γ

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66


From similar triangles

ε εcc st

x d x

=

−

0 0035236 457 236.

=−εst ⇒ ε st = 0.0033

ε

γy

y ms

s

/

/ .

.= =

×=

fE

500 1 15200 10

0 002176

< ε st

Therefore the steel has yielded and the steel stress is 0.87fy as assumed.Lever arm, z, is

z = d − 0.45x = 457 − 0.45 × 236 = 351 mm

Moment capacity, M, is

M =

0 67. fcu

mcγ0.9xbz

M =

0 67 301 5

. .×

0.9 × 236 × 300 × 351 × 10−6 = 299.7 kNm

Maximum uniformly distributed imposed load, qk

Dead loadSelf weight of beam (gk) = 0.5 × 0.3 × 24 = 3.6 kNm−1

Ultimate loadTotal ultimate load (W ) = (1.4gk + 1.6qk)span

= (1.4 × 3.6 + 1.6qk)7

Imposed load


W qB8

5 04 1 6 78

2

( . . )

=+ k = 299.7 kNm (from above)

Hence the maximum uniformly distributed imposed load the beam can support is

qk =

( . )/ ..

299 7 8 7 5 041 6

2× − = 27.4 kNm−1

(B) MAXIMUM POINT LOAD AT MID-SPAN, Q k

g k = 3.6 kN m−1 (from above)

Qk

7 m

qk

gk

7 m

9780415467193_C03a 9/3/09, 12:49 PM66

67

Loading

Ultimate loadUltimate dead load (WD) = 1.4gk × span = 1.4 × 3.6 × 7 = 35.3 kNUltimate imposed load (WI) = 1.6Qk

Imposed loadMaximum design moment, M, is

M

W W = +D I

8 4b b

(Example 2.5, beam B1–B3) =

×+

×

.

. 35 3 78

1 6 74

Qk = 299.7 kNm (from above)

Hence the maximum point load which the beam can support at mid-span is

Q k kN

( . . / )

. =

− ××

=299 7 35 3 7 8 4

1 6 796


Beams

The required compressive strength can beachieved by increasing the proportions of the beam,particularly its overall depth. However, this maynot always be possible due to limitations on theheadroom in the structure, and in such cases it willbe necessary to provide reinforcement in the com-pression face. The compression reinforcement willbe designed to resist the moment in excess of Mu.This will ensure that the compressive stress in theconcrete does not exceed the permissible value andensure an under-reinforced failure mode.

Beams which contain tension and compressionreinforcement are termed doubly reinforced. Theyare generally designed in the same way as singlyreinforced beams except in respect of the calcula-tions needed to determine the areas of tension andcompression reinforcement. This aspect is discussedbelow.

3.9.2.1 Compression and tensile steel areas(clause 3.4.4.4, BS 8110)

The area of compression steel (As′) is calculatedfrom

A

M Mf d d

′su

y

)

. (

=−

− ′0 87(3.18)

where d ′ is the depth of the compression steel fromthe compression face (Fig. 3.33).

The area of tension reinforcement is calculatedfrom

A

M

f zAs

u

ys= +

.

0 87′ (3.19)

where z = d [ . ( . / . )]0 5 0 25 0 9+ − K ′ and K ′ = 0.156.

Equations 3.18 and 3.19 can be derived usingthe stress block shown in Fig. 3.33. This is basic-ally the same stress block used in the analysis of asingly reinforced section (Fig. 3.17) except for theadditional compression force (Fsc) in the steel.

In the derivation of equations 3.18 and 3.19 it isassumed that the compression steel has yielded (i.e.design stress = 0.87fy) and this condition will bemet only if

′ ≤ ′ ≤ =−d

xdd

xd z

. .

0 37 0 19or where 0.45

If d ′/x > 0.37, the compression steel will nothave yielded and, therefore, the compressive stresswill be less than 0.87fy. In such cases, the designstress can be obtained using Fig. 3.9.

Area of concretein compression

Neutral axis

3.9.2 DOUBLY REINFORCED BEAM DESIGNIf the design moment is greater than the ultimatemoment of resistance, i.e. M > Mu, or K > K ′where K = M/fcubd2 and K ′ = Mu/fcubd2 the con-crete will have insufficient strength in compressionto generate this moment and maintain an under-reinforced mode of failure.

9780415467193_C03a 9/3/09, 12:49 PM67


68

d

A ′sd ′

b

x = d/2

ε sc

ε st

Strains

s = 0.9x

Fst

Stress block

Fsc

Fcc

0.45fcu

Section

As

0.0035

zNeutralaxis

Fig. 3.33 Section with compression reinforcement.

DESIGN MOMENT, M

Loading

Ultimate load

Total ultimate load (W) = (1.4gk + 1.6qk)span

= (1.4 × 4 + 1.6 × 5)9 = 122.4 kN

Design moment


Wb8

kNm .

. =×

=122 4 9

8137 7

Example 3.7 Design of bending reinforcement for a doubly reinforcedbeam (BS 8110)

The reinforced concrete beam shown in Fig. 3.34 has an effective span of 9 m and carries uniformly distributed dead(including self weight of beam) and imposed loads of 4 and 5 kNm−1 respectively. Design the bending reinforcementassuming the following:

fcu = 30 Nmm−2

fy = 500 Nmm−2

Cover to main steel = 40 mm

qk = 5 kN m−1

gk = 4 kN m−1h = 370 mm

b = 230 mm

Section A–A

A

A

9 m

Fig. 3.34

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69

2H16

3H25

d ′ = 48

d = 317

Beams

Example 3.7 continuedULTIMATE MOMENT OF RESISTANCE, Mu

Effective depth, dAssume diameter of tension bars (Φ) = 25 mm:

d = h − Φ/2 − cover

= 370 − 25/2 − 40 = 317 mm

Ultimate moment

Mu = 0.156fcubd 2

= 0.156 × 30 × 230 × 3172

= 108.2 × 106 Nmm = 108.2 kNm

Since M > Mu compression reinforcement is required.

COMPRESSION REINFORCEMENTAssume diameter of compression bars (φ) = 16 mm. Hence

d ′ = cover + φ/2 = 40 + 16/2 = 48 mm

z = d [ . ( . / . )]0 5 0 25 0 9+ − ′K = 317 [ . ( . . / . )]0 5 0 25 0 156 0 9+ − = 246 mm

x

d z

.

=−

=−

=0 45

317 2460 45

158.

mm

dx′ =

48158

= 0.3 < 0.37, i.e. compression steel has yielded.

A

M M

f d d′

′)s

u

y

2

0.87mm=

−−

=−

× −=

(

( . . )

. ( )

137 7 108 210

0 87 500 317 48252

6

Hence from Table 3.10, provide 2H16 (A s′ = 402 mm2)

(III) TENSION REINFORCEMENT

A

M

f zAs

u

ys

2

0.87mm

.

. = + =

×× ×

+ =′108 2 10

087 500 246252 1263

6

Hence provide 3H25 (A s = 1470 mm2).

9780415467193_C03a 9/3/09, 12:50 PM69


70

M/b

d2

(N m

m−2

)

14

13

12

11

10

9

8

7

6

5

4

3

2

1

4.0

fcu

0

0.5

1.0

100A

′ s/bd

b

x A′sd

As

d ′

f y

d ′/d 0.15

500

30

2.01.5

x/d = 0.3x/d = 0.4x/d = 0.5

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

0.5 1.0 1.5 2.0 2.5 3.0 3.5

100As/bd

Using the figures given in Example 3.7, Mu =108.2 kNm < M = 137.7 kNm

Since d ′/d (= 48/317) = 0.15 and fcu = 30 N/mm2,chart 7 is appropriate. Furthermore, since the beamis simply supported, no redistribution of momentsis possible, therefore, use x/d = 0.5 constructionline in order to determine areas of reinforcement.

Mbd2

.

.=×

×=

137 7 10230 317

5 956

2

100A s′/bd = 0.33 ⇒ A s′ = 243 mm2

100As/bd = 1.72 ⇒ As = 1254 mm2

Hence from Table 3.10, provide 2H16 compres-sion steel and 3H25 tension steel.

3.9.3 CONTINUOUS, L AND T BEAMSIn most real situations, the beams in buildings areseldom single span but continuous over the sup-ports, e.g. beams 1, 2, 3 and 4 in Fig. 3.36(a). Thedesign process for such beams is similar to thatoutlined above for single span beams. However,the main difference arises from the fact that withcontinuous beams the designer will need to con-sider the various loading arrangements discussedin section 3.6.2 in order to determine the design

3.9.2.2 Design chartsRather than solving equations 3.18 and 3.19 it ispossible to determine the area of tension and com-pression reinforcement simply by using the designcharts for doubly reinforced beams given in Part 3of BS 8110. Such charts are available for designinvolving the use of concrete grades 25, 30, 35,40, 45 and 50 and d ′/d ratios of 0.1, 0.15 and0.2. Unfortunately, as previously mentioned, BSIissued these charts when grade 460 steel was thenorm rather than grade 500 and, therefore, use ofthese charts will overestimate the steel areas byaround 10 per cent. Fig. 3.35 presents a modifiedversion of chart 7 for grade 500 reinforcement.

The design procedure involves the followingsteps:

1. Check Mu < M.2. Calculate d ′/d.3. Select appropriate chart from Part 3 of BS 8110

based on grade of concrete and d ′/d ratio.4. Calculate M/bd 2.5. Plot M/bd 2 ratio on chart and read off corres-

ponding 100A s′/bd and 100A s/bd values (Fig.3.35)

6. Calculate A s′ and As.

Fig. 3.35 Design chart for doubly reinforced beams (based on chart 7, BS 8110: Part 3).

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71

Beams

Table 3.19 Design ultimate moments and shear forces for continuous beams (Table 3.5, BS 8110)

End support End span Penultimate support Interior span Interior support

Moment 0 0.09Fb −0.11Fb 0.07Fb −0.08FbShear 0.45F – 0.6F – 0.55F

F = 1.4Gk + 1.6Qk; b = effective span

Section X–X

T-beam L-beam

Effective flange

X

A

4

X

Effectiveflange

3

2

1

B C D

1 2 3 4

(a)

(b)

Fig. 3.36 Floor slab: (a) plan (b) cross-section.

moments and shear forces in the beam. The analy-sis to calculate the bending moments and shearforces can be carried out by moment distributionas discussed in section 3.9.3.1 or, provided theconditions in clause 3.4.3 of BS 8110 are satisfied(see Example 3.10), by using the coefficients givenin Table 3.5 of BS 8110, reproduced as Table 3.19.Once this has been done, the beam can be sizedand the area of bending reinforcement calculatedas discussed in section 3.9.1 or 3.9.2. At the inter-nal supports, the bending moment is reversed andit should be remembered that the tensile reinforce-ment will occur in the top half of the beam andcompression reinforcement in the bottom half ofthe beam.

Generally, beams and slabs are castmonolithically, that is, they are structurally tied. Atmid-span, it is more economical in such cases to

design the beam as an L or T section by includingthe adjacent areas of the slab (Fig. 3.36(b)). Theactual width of slab that acts together with thebeam is normally termed the effective flange.According to clause 3.4.1.5 of BS 8110, the effec-tive flange width should be taken as the lesser of(a) the actual flange width and (b) the web widthplus bz/5 (for T-beams) or bz/10 (for L-beams),where bz is the distance between points of zeromoments which for a continuous beam may betaken as 0.7 times the distance between the centresof supports.

The depth of the neutral axis in relation to thedepth of flange will influence the design processand must therefore be determined. The depth of theneutral axis, x, can be calculated using equation3.9 derived in section 3.9.1, i.e.

x

d z

=

−0.45

Where the neutral axis lies within the flange,which will normally be the case in practice, thebeam can be designed as being singly reinforcedtaking the breadth of the beam, b, equal to theeffective flange width. At the supports of a con-tinuous member, e.g. at columns B2, B3, C2 andC3, due to the moment reversal, b should be takenas the actual width of the beam.

3.9.3.1 Analysis of continuous beamsContinuous beams (and continuous slabs that spanin one direction) are not statically determinate andmore advanced analytical techniques must be usedto obtain the bending moments and shear forces inthe member. A straightforward method of calculat-ing the moments at the supports of continuousmembers and hence the bending moments andshear forces in the span is by moment distribu-tion. Essentially the moment-distribution methodinvolves the following steps:

1. Calculate the fixed end moments (FEM) in eachspan using the formulae given in Table 3.20 andelsewhere. Note that clockwise moments are con-ventionally positive and anticlockwise momentsare negative.

9780415467193_C03a 9/3/09, 12:51 PM71


72

L

(b)(a)

L

Fig. 3.37 Stiffness factors for uniform beams: (a) pinned-fixed beam = 4EI/L; (b) pinned-pinned beam = 3EI/L.

5. Determine the moment developed at the far endof each member via the carry-over factor. If thefar end of the member is fixed, the carry overfactor is half and a moment of one-half of theapplied moment will develop at the fixed end. Ifthe far end is pinned, the carry over factor iszero and no moment is developed at the far end.

6. Repeat steps (4) and (5) until all the out ofbalance moments are negligible.

7. Determine the end moments for each span bysumming the moments at each joint.

Once the end moments have been determined,it is a simple matter to calculate the bendingmoments and shear forces in individual spans usingstatics as discussed in Chapter 2.

2. Determine the stiffness factor for each span. Thestiffness factor is the moment required to pro-duce unit rotation at the end of the member. Auniform member (i.e. constant EI) of length Lthat is pinned at one end and fixed at the other(Fig. 3.37(a)) has a stiffness factor of 4EI/L. Ifthe member is pinned at both ends its stiffnessfactor reduces to (3/4)4EI/L (Fig. 3.37(b)).

3. Evaluate distribution factors for each membermeeting at a joint. The factors indicate whatproportion of the moment applied to a joint isdistributed to each member attached to it in orderto maintain continuity of slope. Distribution fac-tors are simply ratios of the stiffnesses of indi-vidual members and the sum of the stiffnessesof all the members meeting at a joint. As such,the distribution factors at any joint should sumto unity.

4. Release each joint in turn and distribute theout-of-balance moments between the membersmeeting at the joint in proportion to their dis-tribution factors. The out-of-balance moment isequal in magnitude but opposite in sense to thesum of the moments in the members meeting ata joint.

Example 3.8 Analysis of a two-span continuous beam using momentdistribution

Evaluate the critical moments and shear forces in the beams shown below assuming that they are of constant sectionand the supports provide no restraint to rotation.

W = 100 kN W = 100 kN

B

Load case A

CA

L = 10 m L = 10 m

W = 100 kN W = 100 kN

5 m5 m

L = 10 m L = 10 m

A B C

Load case B

Table 3.20 Fixed end moments for uniformbeams

L/2

ω per unit length

W

8

WL−

MAB MBA

8

WL

L/2

L/2

ωL2

12− ωL2

12

9780415467193_C03a 9/3/09, 12:51 PM72

73

W = 100 kN ≡ 10 kN m−1

MBA = 125 kN m

BAAL = 10 m

Beams


Support reactions and mid-span momentsThe support reactions and mid-span moments are obtained using statics.

LOAD CASE A

Fixed end momentsFrom Table 3.20

MAB = MBC =

−=

− ×= −

.

WL12

100 1012

83 33 kNm

MBA = MCB = =

×=

.

WL12

100 1012

83 33 kNm

Stiffness factorsSince both spans are effectively pinned at both ends, the stiffness factors for members AB and BC (KAB and KBC

respectively), are (3/4)4EI/10.

Distribution factors

Distribution factor at end BA =

+

Stiffness factor for member

Stiffness factor for member Stiffness factor for member AB

AB BC

=

+=

+=

( / ) ( /

.K

K KEI

EI EIAB

AB BC

4 /10(3/4)4 /10 )4 /10

3 43 4

0 5

Similarly the distribution factor at end BC = 0.5

End moments

Joint A B C

End AB BA BC CB

Distribution factors 0.5 0.5FEM (kNm) −83.33 83.33 −83.33 83.33Release A & C1 +83.33 −83.33Carry over2 41.66 −41.66Release B 0

Sum3 (kNm) 0 125 −125 0

Notes:1Since ends A and C are pinned, the moments here must be zero. Applying moments thatare equal in magnitude but opposite in sense to the fixed end moments, i.e. +83.33 kNm and−83.33 kNm, satisfies this condition.2Since joint B is effectively fixed, the carry-over factors for members AB and BC are both0.5 and a moment of one-half of the applied moment will be induced at the fixed end.3Summing the values in each column obtains the support moments.

9780415467193_C03a 9/3/09, 12:51 PM73


74

70.7 kN m

−37.5 kN−62.5 kN

62.5 kN

−125 kNm

A B C

L = 10 m L = 10 m

Example 3.8 continuedTaking moments about end BA obtains the reaction at end A, RA, as follows

10RA = WL/2 − MBA = 100 × (10/2) − 125 = 375 kNm⇒ RA = 37.5 kN

Reaction at end BA, RBA = W − RAB = 100 − 37.5 = 62.5 kN. Since the beam is symmetrically loaded, the reaction atend BC, RBC = RBA. Hence, reaction at support B, RB = RBA + RBC = 62.5 + 62.5 = 125 kN.

The span moments, Mx, are obtained from

Mx = 37.5x − 10x2/2

Maximum moment occurs when ∂M/∂x = 0, i.e. x = 3.75 m ⇒ M = 70.7 kNm. Hence the bending moment and shearforce diagrams are as follows

0.37

5W

1.25

W

0.37

5W

0.07WL 0.07WL

−0.125WL

LOAD CASE B

Fixed end moments

MAB = MBC =

−=

− ×= −

WL8

100 108

125 kNm

MBA = MCB = =

×=

WL8

100 108

125 kNm

The results can be used to obtain moment and reaction coefficients by expressing in terms of W and L, whereW is the load on one span only, i.e. 100 kN and L is the length of one span, i.e. 10 m, as shown in Fig. 3.38. Thecoefficients enable the bending moments and shear forces of any two equal span continuous beam, subjected touniformly distributed loading, to be rapidly assessed.

Fig. 3.38 Bending moment and reaction coefficients for two equal span continuous beams subjected to a uniform load of W oneach span.

9780415467193_C03a 9/3/09, 12:52 PM74

75

Beams

Example 3.8 continuedStiffness and distribution factorsThe stiffness and distribution factors are unchanged from the values calculated above.

End moments

Support reactions and mid-span momentsThe support reactions and mid-span moments are again obtained using statics.

Fig. 3.39

By taking moments about end BA, the reaction at end A, RA, is

10RA = WL/2 − MBA = 100 × (10/2) − 187.5 = 312.5 kNm ⇒ RA = 31.25 kN

The reaction at end BA, RBA = W − RA = 100 − 31.25 = 68.75 kN = RBC. Hence the total reaction at support B,RB = RBA + RBC = 68.75 + 68.75 = 137.5 kN

By inspection, the maximum sagging moment occurs at the point load, i.e. x = 5 m, and is given by

Mx=5 = 31.25x = 31.25 × 5 = 156.25 kNm

The bending moments and shear forces in the beam are shown in Fig. 3.39. Fig. 3.40 records the moment and reactioncoefficients for the beam.

Joint A B C

End AB BA BC CB

Distribution factors 0.5 0.5FEM (kNm) −125 125 −125 125Release A & C +125 −125Carry over 62.5 −62.5Release B 0

Sum 0 187.5 −187.5 0

31.25 kN

−68.75 kN

68.75 kN

−31.25 kN

156.25 kN m

−187.5 kN m

A B C

W = 100 kN W = 100 kN

L = 10 mA BA

W = 100 kN

MBA = 187.5 kN m

9780415467193_C03a 9/3/09, 12:53 PM75


76


Fig. 3.40 Bending moment and reaction coefficients for a two equal span continuous beam subjected to concentrated loads ofW at each mid-span.

0.31

25W

1.37

5W

0.31

25W

0.156WL

W

0.156WL

W

−0.188WL

Example 3.9 Analysis of a three span continuous beam using momentdistribution

A three span continuous beam of constant section on simple supports is subjected to the uniformly distributed loadsshown below. Evaluate the critical bending moments and shear forces in the beam using moment distribution.

L = 10 mL = 10 mL = 10 m

A B C D

W = 100 kN W = 100 kNW = 100 kN � 10 kN/m

FIXED END MOMENTS (FEM)

MAB = MBC = MCD =

−=

− ×= −

WL12

100 1012

83 33

. kN m

MBA = MCB = MDC =

WL12

100 1012

83 33

. =×

= kN m

STIFFNESS FACTORSThe outer spans are effectively pinned at both ends. Therefore, the stiffness factors for members AB and CD (KAB andKCD respectively), are (3/4)4EI/10.

During analysis, span BC is effectively pinned at one end and fixed at the other and its stiffness factor, KBC, istherefore 4EI/10.

DISTRIBUTION FACTORS

Distribution factor at end BA =

Stiffness factor for member Stiffness factor for member Stiffness factor for member

ABAB BC+

=

+=

+=

( / )

K

K KEI

EI EIAB

AB BC

4 /10(3/4)4 /10 4 /10

3 4 37

Distribution factor at end BC =

+

Stiffness factor for member

Stiffness factor for member Stiffness factor for member BC

AB BC

=

+=

+=

/ /

KK K

EIEI EI

AB

AB BC

4 /10(3/4)4 10 4 10

47

Similarly the distribution factors for ends CB and CD are, respectively, 4/7 and 3/7.

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77

END MOMENTS

Beams

W = 100 kN ≡ 10 kN/m

L = 10 mA BA

MBA = 100 kN m

Taking moments about end BA obtains the reaction at end A, RA, as follows

10RA = WL/2 − MBA = 100 × (10/2) − 100 = 400 kNm ⇒ RA = 40 kN

Reaction at end BA, RBA = W − RAB = 100 − 40 = 60 kNThe span moments, Mx, are obtained from

Mx = 40x − 10x2/2

Maximum moment occurs when ∂M/∂x = 0, i.e. x = 4.0 m ⇒ M = 80 kNm.

Span BCW = 100 kN ≡ 10 kN/m

L = 10 mBC

MBA = 100 kN m

CB

MBA = 100 kN m


Joint A B C D

End AB BA BC CB CD DC

Distribution factors 3/7 4/7 4/7 3/7FEM (kNm) −83.33 83.33 −83.33 83.33 −83.33 83.33Release A & D +83.33 −83.33Carry over 41.66 −41.66Release B & C −17.86 −23.8 23.8 17.86Carry over 11.9 −11.9Release B & C −5.1 −6.8 6.8 5.1Carry over 3.4 −3.4Release B & C −1.46 −1.94 1.94 1.46Carry over 0.97 −0.97Release B & C −0.42 −0.55 0.55 0.42Carry over 0.28 −0.28Release B & C −0.12 −0.16 0.16 0.12

Sum 0 100.04 −100.04 100.04 −100.04 0

SUPPORT REACTIONS AND MID-SPAN MOMENTS

Span AB

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78

Example 3.9 continuedBy inspection, reaction at end BC, RBC, = reaction at end CB, RCB = 50 kNTherefore, total reaction at support B, RB = RBA + RBC = 60 + 50 = 110 kNBy inspection, maximum moment occurs at mid-span of beam, i.e. x = 5 m. Hence, maximum moment is given by

Mx=5 = 50x − 10x2/2 − 100 = 50 × 5 − 10 × 52/2 − 100 = 25 kNm

The bending moment and shear force diagrams plus the moment and reaction coefficients for the beam are shownbelow.

L = 10 m L = 10 m

DCBA

−100 kN m

25 kN m

60 kN80 kN m

−50 kN −40 kN

0.08WL 0.025WL 0.08WL

−0.1WL −0.1WL

0.4W

1.1W

1.1W

0.4W

L = 10 m

150

400

300

3750

Example 3.10 Continuous beam design (BS 8110)A typical floor plan of a small building structure is shown in Fig. 3.41. Design continuous beams 3A/D and B1/5 assum-ing the slab supports an imposed load of 4 kNm−2 and finishes of 1.5 kNm−2. The overall sizes of the beams and slab areindicated on the drawing. The columns are 400 × 400 mm. The characteristic strength of the concrete is 35 Nmm−2

and of the steel reinforcement is 500 Nmm−2. The cover to all reinforcement may be assumed to be 30 mm.

GRID LINE 3

Loading

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79

Beams


5

4

3

2

1

675

× 40

0

150

550 × 300

3.75 m

3.75 m

3.75 m

3.75 m

A B C D

8.5 m 8.5 m 8.5 m

Fig. 3.41

Dead load, gk, is the sum of

weight of slab = 0.15 × 3.75 × 24 = 13.5weight of downstand = 0.3 × 0.4 × 24 = 2.88finishes = 1.5 × 3.75 = 5.625

22.0 kNm−1

Imposed load, qk = 4 × 3.75 = 15 kNm−1

Design uniformly distributed load, ω = (1.4gk + 1.6qk) = (1.4 × 22 + 1.6 × 15) = 54.8 kNm−1

Design load per span, F = ω × span = 54.8 × 8.5 = 465.8 kN

Design moments and shear forcesFrom clause 3.4.3 of BS 8110, as gk > qk, the loading on the beam is substantially uniformly distributed and the spansare of equal length, the coefficients in Table 3.19 can be used to calculate the design ultimate moments and shearforces. The results are shown in the table below. It should be noted however that these values are conservativeestimates of the true in-span design moments and shear forces since the coefficients in Table 3.19 are based onsimple supports at the ends of the beam. In reality, beam 3A/D is part of a monolithic frame and significant restraintmoments will occur at end supports.

9780415467193_C03a 9/3/09, 12:54 PM79


80

51300

4H25

499


Position Bending moment Shear force

Support 3A 0 0.45 × 465.8 = 209.6 kNNear middle of 3A/B 0.09 × 465.8 × 8.5 = 356.3 kNm 0

0.6 × 465.8 = 279.5 kN (support 3B/A*)Support 3B −0.11 × 465.8 × 8.5 = −435.5 kNm

0.55 × 465.8 = 256.2 kN (support 3B/C**)Middle of 3B/C 0.07 × 465.8 × 8.5 = 277.2 kNm 0

* shear force at support 3B towards A ** shear force at support 3B towards C

Steel reinforcement

Middle of 3A/B (and middle of 3C/D)Assume diameter of main steel, φ = 25 mm, diameter of links, φ′ = 8 mm and nominal cover, c = 30 mm. Hence

Effective depth, d = h −

φ2

− φ′ − c = 550 −

252

− 8 − 30 = 499 mm

The effective width of beam is the lesser of

(a) actual flange width = 3750 mm(b) web width + bz/5, where bz is the distance between points of zero moments which for a continuous beam may be

taken as 0.7 times the distance between centres of supports. Hence

bz = 0.7 × 8500 = 5950 mm and b = 300 + 5950/5 = 1490 mm (critical)

K

Mf bd

.

.= =

×× ×

=cu

2

6

2

356 3 1035 1490 499

0 0274

z d K ( ( . )= + −0.5 /0.9)0 25 ≤ 0.95d = 0.95 × 499 = 474 mm (critical)

= + − ( . ( . . / . ))499 0 5 0 25 0 0274 0 9 = 499 × 0.969 = 483 mm

x = (d − z)/0.45 = (499 − 474)/0.45 = 56 mm < flange thickness (= 150 mm). Hence

Area of steel reinforcement, A s =

M

f z0.87mm

y

2 .

. ( . ) =

×× ×

=356 3 10

0 87 500 0 95 4991728

6

= 1728 mm2

Provide 4H25 (A s = 1960 mm2).

9780415467193_C03a 9/3/09, 12:55 PM80

81

63

436

51 300

2H25

6H25

At support 3B (and 3C)Assume the main steel consists of two layers of 25 mm diameter bars, diameter of links, φ′ = 8 mm and nominalcover, c = 30 mm. Hence

Effective depth, d = h − φ − φ′ − c = 550 − 25 − 8 − 30 = 487 mm

Since the beam is in hogging, b = 300 mm

Mu = 0.156fcubd2 = 0.156 × 35 × 300 × 4872 × 10−6 = 388.5 kNm

Since Mu < M (= 435.5 kNm), compression reinforcement is required.Assume diameter of compression steel, Φ = 25 mm, diameter of links, φ′ = 8 mm, and cover to reinforcement, c, is30 mm. Hence effective depth of compression steel d ′ is

d ′ = c + φ′ + Φ/2 = 30 + 8 + 25/2 = 51 mm

Lever arm, z d K ( . ( . / . )) ( . ( . . / . )) = + − = + − =0 5 0 25 0 9 487 0 5 0 25 0 156 0 9 378′ mm

Depth to neutral axis, x = (d − z)/0.45 = (487 − 378)/0.45 = 242 mmd′/x = 51/242 = 0.21 < 0.37. Therefore, the compression steel has yielded, i.e. f s′ = 0.87fy and

Area of compression steel, A

M M

f d ds

u

y

2

0.87 (mm′

′)=

−−

=−

× −=

( . . )

. ( )

435 5 388 510

0 87 500 487 51248

6

Provide 2H25 (A s′ = 982 mm2)

Area of tension steel, A

M

f zAs

u

ys

2

0.87mm

.

. = + =

×× ×

+ =′388 5 10

0 87 500 378248 2610

6

Provide 6H25 as shown (A s = 2950 mm2)

Beams


Middle of 3B/CFrom above, effective depth, d = 499 mm and effective width of beam, b = 1490 mm.Hence, A s is

A

M

f zs

y

2

0.87mm

.

. ( . ) = =

×× ×

=277 2 10

0 87 500 0 95 4991344

6

Provide 3H25 (A s = 1470 mm2).Fig. 3.42 shows a sketch of the bending reinforcement for spans 3A/B and 3B/C. The curtailment lengths indicated onthe sketch are in accordance with the simplified rules for beams given in clause 3.12.10.2 of BS 8110 (Fig. 3.27).

9780415467193_C03a 9/3/09, 12:56 PM81


82

3A 3B 3C

850

x1

2H254H25

8100

3H25

8100400 400

1275 2H25

2H25

x31minimum linksx31x21minimum links

x3

4H252H25 2H254H25

2H252125 1275

x3x2


ShearAs discussed in section 3.9.1.3, the amount and spacing of shear reinforcement depends on the area of tensile steelreinforcement present in the beam. Near to supports and at mid-spans this is relatively easy to asses. However, atintervening positions on the beam this task is more difficult because the points of zero bending moment areunknown. It is normal practice therefore to use conservative estimates of A s to design the shear reinforcementwithout obtaining detailed knowledge of the bending moment distribution as illustrated below.

Span 3A/B (3B/C and 3C/D)The minimum tension steel at any point in the span is 2H25. Hence A s = 980 mm2 and

100 100 980300 499

0 655A

bds

.=××

=

From Table 3.11, νc =

3525

3

× 0.54 = 0.6 Nmm−2

Provide minimum links where V ≤ (νc + 0.4)bd = (0.6 + 0.4)300 × 499 × 10−3 = 149.7 kN (Fig. 3.42), according to

A

sbf

s

y0.87ν

ν ν

.

. .

.= =××

=0 4 0 4 300

0 87 5000 276with s v ≤ 0.75d = 373 mm

From Table 3.13 select H8 at 300 centres (A sv/s v = 0.335)

Support 3A (and 3D)According to clause 3.4.5.10 of BS 8110 for beams carrying generally uniform load the critical section for shear maybe taken at distance d beyond the face of the support, i.e. 0.2 + 0.499 = 0.7 m from the column centreline. Here thedesign shear force, VD, is

VD = V3A − 0.7ω = 209.6 − 0.7 × 54.8 = 171.3 kN

Shear stress, ν =

V

bdD 2Nmm

.

. =

××

= −171 3 10

499 3001 14

3

Fig. 3.42

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83

Beams

Example 3.10 continuedFrom Fig. 3.42 it can be seen that 50 per cent of main steel is curtailed at support A. The effective area of tension

steel is 2H25, hence A s = 980 mm2,

100Abd

s = 0.655 and νc = 0.6 Nmm−2

Since ν > (νc + 0.4) provide design links according to

A

sb

fs c

y

(ν

ν ν

ν ν

).

( . . ).

. .≥−

=−

×=

0 87300 1 14 0 6

0 87 5000 372

From Table 3.13 select H8 links at 250 mm centres (A sv/s v = 0.402).Note that the shear resistance obtained with minimum links, V1, is

V1 = (νc + 0.4)bd = (0.6 + 0.4)300 × 499 × 10−3 = 149.7 kN

This shear force occurs at x1 = (V3A − V1)/ω = (209.6 − 149.7)/54.8 = 1.09 m (Fig. 3.42). Assuming the first link isfixed 75 mm from the front face of column 3A (i.e. 275 mm from the centre line of the column), then five H8 linksat 250 mm centres are required.

Support 3B/A (and 3C/D)Design shear force at distance d beyond the face of the support (= 0.2 + 0.487 = 0.687 m), VD, is

VD = V3B/A − 0.687ω = 279.5 − 0.687 × 54.8 = 241.9 kN

Shear stress, ν =

V

bdD 2Nmm

.

. =

××

= −241 9 10

300 4871 66

3

Assume the tension steel is 4H25. Hence A s = 1960 mm2 and

100 100 1960300 487

1 34A

bds

.=××

=


3525

3

× 0.69 = 0.77 Nmm−2


A

sb

fs c

y

(ν

ν ν

ν ν

).

( . . ).

. .≥−

=−

×=

0 87300 1 66 0 77

0 87 5000 614 Select H8 at 150 centres (A sv/s v = 0.671).

To determine the number of links required assume H8 links at 200 mm centres (A sv/s v = 0.503) are to be providedbetween the minimum links in span 3A/B (i.e. H8@300) and the design links at support 3B (H8@150). In this length,the minimum tension steel is 2H25 and from above νc = 0.6 Nmm−2. The shear resistance of H8 links at 200 mmcentres plus concrete, V2, is

V

A

sf b d2

sy c=

+

. ν

νν0 87

= (0.503 × 0.87 × 500 + 300 × 0.6)499 × 10−3 = 199 kN

Assuming the first link is fixed 75 mm from the front face of column 3B then nine H8 links at 150 mm centres arerequired. The distance from the centre line of column 3B to the ninth link (x2) is (200 + 75) + 8 × 150 = 1475 mm.Since x2 < 2125 mm the tension steel is 4H25 as assumed. The shear force at x2 is

Vx2 = 279.5 − 1.475 × 54.8 = 198.7 kN < V2 OK

From above shear resistance of minimum links = 149.7 kN. This occurs at x21 = (279.5 − 149.7)/54.8 = 2.368 m(Fig. 3.42). Therefore provide five H8 links at 200 mm centres and fifteen H8 links at 300 centres arranged as shownin Fig. 3.43.

9780415467193_C03a 9/3/09, 12:57 PM83


84

5H8@

250

5H8@

200

9H8@

150

8H8@

150

4H8@

225

4H8@

225

8H8@

150

15H8@300 13H8@300

275 2H25200

4H25225

2H25

225

3H252H2512754H252H25

850

40081008100

275

275

400

3A 3B 3C

Example 3.10 continued(d) Support 3B/C (and 3C/B)Shear force at distance d from support 3B/C, VD, is

VD = V3B/A − 0.687ω = 256.2 − 0.687 × 54.8 = 218.6 kN

Shear stress, νc =

Vbd

D .

=×

×218 6 10300 487

3

= 1.50 Nmm−2

Again, assume the tension steel is 4H25. Hence A s = 1960 mm2,

100Abd

s = 1.34, and νc = 0.77 Nmm−2.


A

sb

fs c

y

(ν

ν ν

ν ν

).

( . . ).

.≥−

=−

×=

0 87300 1 50 0 77

0 87 5000 503. Select H8 at 150 centres (A sv/s v = 0.671).

To determine the number of links required assume H8 links at 225 mm centres (A sv/sv = 0.447) are to be providedbetween the minimum links in span 3B/C and the design links at support 3B. In this length the minimum tension steelis 2H25 and from above νc = 0.6 Nmm−2. The shear resistance of H8 links at 225 mm centres plus concrete, V3, is

V3

A

sf b ds

y c=

+

. ν

νν0 87

= (0.447 × 0.87 × 500 + 300 × 0.6)499 × 10−3 = 186.8 kN

Assuming the first link is fixed 75 mm from the front face of column 3B then eight H8 links at 150 mm centres arerequired. The distance from the centre line of column 3B to the eight link (x3) is (200 + 75) + 7 × 150 = 1325 mm.Since x3 < 2125 mm the tension steel is 4H25 as assumed. The shear force at x3 is

Vx3 = 256.2 − 1.325 × 54.8 = 183.6 kN < V3 OK

From above shear resistance of minimum links = 149.7 kN. This occurs at x31 = (256.2 − 149.7)/54.8 = 1.943 m(Fig. 3.42). Therefore provide four H8 links at 225 mm centres and thirteen H8 links at 300 centres arranged as shownin Fig. 3.43.

Fig. 3.43 shows the main reinforcement requirements for spans 3A/B and 3B/C. Note that in the above calculationsthe serviceability limit state of cracking has not been considered. For this reason as well as reasons of buildab-ility, the actual reinforcement details may well be slightly different to those indicated in the figure. As previouslynoted the beam will be hogging at end supports and further calculations will be necessary to determine the areaof steel required in the top face.

Fig. 3.43

9780415467193_C03a 11/3/09, 11:04 AM84

85

Beams

Example 3.10 continuedDeflection

Actual

span

effective depth = =

8500

49917

By inspection, exterior span is critical.

bbw =

3001490

= 0.2 < 0.3 ⇒ basic span/effective depth ratio of beam = 20.8 (Table 3.12)

Service stress, fs, is

f f

A

As y

s,req

s,pro

2Nmm= = × × = − 5

8

5

8500

1728

1960276

ν

modification factor

= +−

+

= +−

+×

×

= .

.

. .

.0 55447

0 55477 276

120 0 9356 3 101490 499

1 456

2

fM

bd

s

2120 0.9

Therefore,

permissible

span

effective depth = basic ratio × mod.factor = 20.8 × 1.45 = 30 > actual OK

PRIMARY BEAM ON GRID LINE B

B1 B2 B3 B4 B5

L = 7.5 m L = 7.5 m

R2B/A + R2B/C R4B/A + R4B/C

Self-weight of downstand

Fig. 3.44

675

400

150

LoadingDesign load on beam = uniformly distributed load from self weight of downstand

+ reactions at B2 and B4 from beams 2A/B, 2B/C, 4A/B and 4B/C,i.e. R2B/A, R2B/C, R4B/A and R4B/C.

Uniform loadsDead load from self weight of downstand, gk = 0.4 × (0.675 − 0.15) × 24 = 5.04 kNm−1

≡ 5.04 × 7.5 = 37.8 kN on each span.

Imposed load, qk = 0

Point loadsDead load from reactions R2B/A and R2B/C (and R4B/A and R4B/C), Gk, is

Gk = 22 × (0.6 × 8.5) + 22 × 4.25 = 205.7 kN

9780415467193_C03a 9/3/09, 12:58 PM85


86

Example 3.10 continuedImposed load from reactions R2B/A and R2B/C (and R4B/A and R4B/C), Q k, is

Q k = 15 × (0.6 × 8.5) + 15 × 4.25 = 140.3 kN

Load casesSince the beam does not satisfy the conditions in clause 3.4.3, the coefficients in Table 3.19 cannot be used toestimate the design moments and shear forces. They can be obtained using techniques such as moment distribution,as discussed in section 3.9.3.1. As noted in section 3.6.2 for continuous beams, two load cases must be considered:(1) maximum design load on all spans (Fig. 3.45(a)) and (2) maximum and minimum design loads on alternate spans(Fig. 3.45(b)). Assume for the sake of simplicity that the ends of beam B1/5 are simple supports.

Maximum design load = uniform load (W ′ = 1.4gk + 1.6qk = 1.4 × 37.8 + 0 = 52.9 kN)+ point load (W ″ = 1.4Gk + 1.6Qk = 1.4 × 205.7 + 1.6 × 140.3 = 512.5 kN)

Minimum design load = uniform load (W ′″ = 1.0gk = 1.0 × 37.8 = 37.8 kN)+ point load (W ″″ = 1.0Gk = 1.0 × 205.7 = 205.7 kN)

Load case 1

Fixed end momentsFrom Table 3.20

MB1/3 = MB3/5 = − − = −

×−

×= −

W L W L′ ″12 8

kN m . .

. .

. 52 9 7 5

12512 5 7 5

8513 5

MB3/1 = MB5/3 =

W L W L′ ″12 8

kN m . + = 513 5

Stiffness and distribution factorsReferring to Example 3.8 it can be seen that the stiffness factors for members B3/1 and B3/5 are both (3/4)4EI/7.5.Therefore the distribution factor at end B3/1 and end B3/5 are both 0.5.

B1 B3 B5

52.9 kN 37.8 kN

205.7 kN512.5 kN

B1 B3 B5

52.9 kN 52.9 kN

512.5 kN512.5 kN

(a)

(b)

Fig. 3.45 Load cases: (a) load case 1 (b) load case 2.

9780415467193_C03a 9/3/09, 12:59 PM86

87

Beams

Example 3.10 continuedEnd moments

Joint B1 B3 B5

End B1/3 B3/1 B3/5 B5/3

Distribution factors 0.5 0.5FEM (kNm) −513.5 513.5 −513.5 513.5Release B1 & B5 +513.5 −513.5Carry over 256.8 −256.8

Sum (kNm) 0 770.3 −770.3 0

Span moments and reactionsThe support reactions and mid-span moments are obtained using statics.

B1/3 B3/1L = 7.5 m

W = 52.9 kN512.5 kN

770.3 kN m

Taking moments about end B3/1 obtains the reaction at end B1/3, RB1/3, as follows

7.5RB1/3 = 512.5 × (7.5/2) + 52.9 × (7.5/2) − 770.3 = 1350 kNm ⇒ RB1/3 = 180 kN

Reaction at end B3/1, RB3/1 = 512.5 + 52.9 − 180 = 385.4 kN. Since the beam is symmetrically loaded, the reactionat end B3/5 is 385.4 kN and end B5/3 is 180 kN.

By inspection, the maximum sagging moment occurs at the point load, i.e. x = 3.75 m, and is given by

Mx=3.75m = 180x − (52.9/7.5)x2/2 = 625.4 kNm

Load case 2

Fixed end moments

M M

W L W LB1/3 B3/1 12 8

kNm= − = − − = −×

−×

= − . .

. .

. ′ ″ 52 9 7 5

12512 5 7 5

8513 5

M M

W L W LB3/5 B5/3 12 8

kNm= − = − − = −×

−×

= − . .

. .

. ′″ ″″ 37 8 7 5

12205 7 7 5

8216 5

Stiffness and distribution factorsThe stiffness and distribution factors are unchanged from the values calculated above.

9780415467193_C03a 9/3/09, 12:59 PM87


88

Example 3.10 continuedEnd moments

B3/1B1/3

52.9 kN 512.5 kN547.6 kN m

L = 7.5 m

B3/5

W = 205.7

L = 7.5 m

547.6 kN m37.8 kN

B5/3

Span moments and reactionsThe support reactions and mid-span moments are obtained using statics.

Joint B1 B3 B5

End B1/3 B3/1 B3/5 B5/3

Distribution factors 0.5 0.5FEM (kNm) −513.5 513.5 −216.5 216.5Release B1 & B5 +513.5 −216.5Carry over 256.8 −108.3Release B3 −222.8* −222.8

Sum (kNm) 0 547.5 −547.6 0

* 0.5 [−(513.5 − 216.5 + 256.8 − 108.3)]

Taking moments about end B3/1 of beam B1/3, the reaction at end B1/3, RB1/3, is

7.5RB1/3 = 512.5 × (7.5/2) + 52.9 × (7.5/2) − 547.6 = 1572.7 kNm

RB1/3 = 209.7 kN

Hence, reaction at end B3/1, RB3/1 = 512.5 + 52.9 − 209.7 = 355.7 kN.Similarly for beam B3/5, the reaction at end B5/3, RB5/3, is

7.5RB5/3 = 205.7 × (7.5/2) + 37.8 × (7.5/2) − 547.6 = 365.5 kNm

RB5/3 = 48.7 kN

and RB3/5 = 205.7 + 37.8 − 48.7 = 194.8 kN

By inspection, the maximum sagging moment in span B1/3 occurs at the point load and is given by

Mx=3.75 = 209.7x − (52.9/7.5)x2/2 = 736.8 kNm

Similarly, the maximum sagging moment in span B3/5 is

Mx=3.75 = 194.8x − (37.8/7.5)x2/2 − 547.6 = 147.5 kNm

9780415467193_C03a 9/3/09, 1:00 PM88

89

610

65400

6H25

Beams

Example 3.10 continuedDesign moments and shear forcesThe bending moment and shear force envelops for load cases 1 and 2 are shown below. It can be seen that the designsagging moment is 736.8 kNm and the design hogging moment is 770.3 kNm. The design shear force at supports B1and B5 is 209.7 kN and at support B3 is 385.4 kN.

Steel reinforcement

Middle of span B1/3 (and B3/5)Assume the main steel consists of two layers of 25 mm diameter bars, diameter of links, φ′ = 10 mm and nominalcover, c = 30 mm. Hence effective depth, d, is

d = h − (φ + φ′ + c) = 675 − (25 + 10 + 30) = 610 mm

From clause 3.4.1.5 of BS 8110 the effective width of beam, b, is

b = bw + 0.7b/5 = 400 + 0.7 × 7500/5 = 1450 mm

K

Mf bd

.

.= =

×× ×

=cu

2

736 8 1035 1450 610

0 0396

2

z d K ( . ( . . ))= + −0 5 0 25 0 9/ ≤ 0.95d = 0.95 × 610 = 580 mm (critical)

= + − ( . ( . . . )6 0 0 / )1 0 5 0 25 039 0 9 = 610 × 0.955 = 583 mm

x = (d − z)/0.45 = (610 − 580)/0.45 = 67 mm < flange thickness (= 150 mm). Hence

Area of tension steel, A

M

f zs

y

2

0.87mm

.

. ( . ) = =

×× ×

=736 8 10

0 87 500 0 95 6102923

6


B1 B3 B5

−770.3 kN mLoad case 2

Load case 1

736.8 kN 385.4 kN

−385.4 kN

209.7 kN

9780415467193_C03a 9/3/09, 1:00 PM89


90

65

610

400

2

8H25

1

Example 3.10 continuedAt support B3Again, assuming that the main steel consists of two layers of 25 mm diameter bars, diameter of links, φ′ = 10 mm andnominal cover, c = 30 mm, implies effective depth, d = 610 mm

Since the beam is in hogging, effective width of beam, b = bw = 400 mm

K

Mf bd

.

.= =

×× ×

=cu

2

770 3 1035 400 610

0 1486

2

z d K ( . ( . . )) ( . ( . . / . )) = + − = + − =0 5 0 25 0 9 610 0 5 0 25 0 148 0 9 483/ mm

A

M

f zs

y

2

0.87mm

.

. = =

×× ×

=770 3 10

0 87 500 4833666

6


Note that, in practice, it would be difficult to hold bars 1 in place and a spacer bar would be needed between the twolayers of reinforcement. This would reduce the value of the effective depth, but this aspect has been ignored in thecalculations.

The simplified rules for curtailment of bars in continuous beams illustrated in Fig. 3.27 do not apply as the loadingon the beam is not predominantly uniformly distributed. However, the general rule given in clause 3.12.9.1 can beused as discussed below with reference to bar marks 1 and 2.

The theoretical position along the beam where mark 1 bars (i.e. 2H25 from the inner layer of reinforcement) canbe stopped off is where the moment of resistance of the section considering only the continuing bars (see sketch), Mr,is equal to the design moment, M.

For equilibrium Fcc = Fst

0 67. fcu

mcγ0.9xb = 0.87fyA s

0 67 351 5

. .×

0.9 × x × 400 = 0.87 × 500 × 2950

Hence x = 228 mm

Also, z = d − 0.9x/2 = 610 − 0.9 × 228/2 = 507 mm

Moment of resistance of section, Mr, is

M F z

fxb zr cc

cu

mc

= =

=×

× ×

.

. .

..

0 670 9

0 67 351 5

0 9 229 400γ

505 × 10−6 = 650 kNm

x

Fst

Fcc

9780415467193_C03a 9/3/09, 1:00 PM90

91

B2

400

B1B1B1 B3 B4 B5

300

2H252H25

300

1200 2H25 2100 2H25

4H25

40013007100

4H25

4H25

2H25

4H25

4800

3600

2600

7100

2H25

The design moment along the beam from end B3/1, M, is

Beams


W = 52.9 kN≡ 7.05 kN/m

512.5 kN

M 770.3 kN m

RB3/1 = 385.4 kNB1/3x

M = 770.3 − 385.4x +

7 052

2. x. Solving for M = 651 kNm implies that the theoretical cut-off point of mark 1 bars

from the centre-line of support B3 is 0.31 m.According to clause 3.12.9.1, the actual cut-off point of bars in the tension zone is obtained by extending the bars

an anchorage length (= 38φ from Table 3.27 of BS 8110 assuming fcu = 35 Nmm−2 and fy = 500 nmm−2, deformed type2 bars) beyond the theoretical cut-off point (i.e. 310 mm + 38 × 25 = 1260 mm) or where other bars continuing pastprovide double the area required to resist the moment at the section i.e. where the design moment is 1/2M = 325.5kNm. The latter is obtained by solving the above expression for x assuming M = 325.5 kNm. This implies that theactual cut-off point of the bars is 1.17 m. Hence the 2H25 bars can be stopped off at, say, 1.3 m from support B3.

The cut-off point of mark 2 bars can be similarly evaluated. Here A s is 1960 mm2. Hence x = 152 mm, z = 552 mmassuming d = 620 mm and Mr = 471 kNm. The theoretical cut-off point of the bars is 0.78 m from the centre-lineof support B3. The actual cut-off point is then either 780 + 38 × 25 = 1730 mm or where the design moment is235.5 kNm, i.e. 1.406 m. Thus it can be assumed the mark 2 bars can be stopped off at, say, 1.8 m from the centre-line of support B3.

Repeating the above procedure will obtain the cut-off points of the remaining sets of bars. Not all bars will needto be curtailed however. Some should continue through to supports as recommended in the simplified rules forcurtailment of reinforcement for beams. Also the anchorage length of bars that continue to end supports or arestopped off in the compression zone will vary and the reader is referred to the provisions in clause 3.12.9.1 for furtherdetails. Fig. 3.46 shows a sketch of the bending reinforcement for the beam.

Fig. 3.46

9780415467193_C03a 9/3/09, 1:01 PM91


92

Example 3.10 continuedShear

Support B1(and B5)

Shear stress, ν

.

. . . . = =

××

= < = =− −V

bdf

209 7 10

400 6200 85 0 8 0 8 35 4 7

3

Nmm Nmm2cu

2 OK

From Fig. 3.46 it can be seen that the area of tension steel is 4H25. Hence A s = 1960 mm2 and

100 100 1960400 620

0 79A

bds

.=××

=

From Table 3.11, νc

2Nmm= × = − . . 35

250 58 0 65

3

Since ν < (νc + 0.4) minimum links are required according to

As

bf

s

y0.87ν

ν ν

.

. .

.= =××

=0 4 0 4 400

0 87 5000 368

Provide H10-300 (A sv/s v = 0.523)

Support B3

Shear stress, ν

.

. . . . = =

××

= < = =− −V

bdf

385 4 10

400 6201 55 0 8 0 8 35 4 7

3

Nmm Nmm2cu

2 OK

From Fig. 3.46 it can be seen that the minimum tension steel at any point between B3 and B2 is 2H25. Hence A s =980 mm2 and

100 100 980400 620

0 40A

bds

.=××

=


3525

3

× 0.46 = 0.52 Nmm−2


As

bf

s c

y0.87ν

ν ν

ν ν

( )

( . . ).

.≥−

=−

×=

400 1 55 0 520 87 500

0 947

Provide H10–150 (A sv/s v = 1.047)Fig. 3.47 shows the main reinforcement requirements for beam B1/5. Note that in the above calculations theserviceability limit state of cracking has not been considered. For this reason as well as reasons of buildability, theactual reinforcement details may well be slightly different to those shown. As previously noted, the beam will behogging at end supports and further calculations will be necessary to determine the area of steel required in the topface.

Deflection

Actual

span

effective depth .= =

7500

61012 3

bbw =

4001450

= 0.276 < 0.3 ⇒ basic span/effective depth ratio of beam = 20.8 (Table 3.12)

9780415467193_C03a 9/3/09, 1:01 PM92

93

400

B1 B2 B3 B4 B5

12H10-300 23H10-150 23H10-150 12H10-300

2H25

4502H25

450

2600

3600

48004H25

2H25

4H25

1200 2H25 2100 2H25

4H254H25

7100 71004001300

275

Service stress, fs, is

f f

A

As y

s,req

s,pro

2Nmm= = × × = − 5

8

5

8500

2923

2950310

ν

Modification factor

= +−

+

= +−

+×

×

= .

. .

. .

.0 55477

120 0 90 55

477 310

120 0 9736 8 101450 610

1 2

2

6

2

fM

bd

s

Therefore, permissible

spaneffective depth

= basic ratio × modification factor

= 20.8 × 1.2 = 25 > actual OK

Slabs


Fig. 3.47

3.9.4 SUMMARY FOR BEAM DESIGNFigure 3.48 shows the basic steps that should befollowed in order to design reinforced concretebeams.

3.10 SlabsIf a series of very wide, shallow rectangular beamswere placed side by side and connected transverselysuch that it was possible to share the load betweenadjacent beams, the combination of beams wouldact as a slab (Fig. 3.49).

Reinforced concrete slabs are used to form avariety of elements in building structures such asfloors, roofs, staircases, foundations and some typesof walls (Fig. 3.50). Since these elements can bemodelled as a set of transversely connected beams,it follows that the design of slabs is similar, inprinciple, to that for beams. The major differenceis that in slab design the serviceability limit state ofdeflection is normally critical, rather than the ulti-mate limit states of bending and shear.

9780415467193_C03a 9/3/09, 1:01 PM93


94

Select:Concrete strength class (say C28/35)Longitudinal reinforcement grade (say 500 Nmm−2)Shear reinforcement grade (say 500 Nmm−2)Minimum member size (see Fig. 3.11)Thickness of concrete cover (see 3.8)

Estimate:Characteristic dead loadCharacteristic imposed loadself weight

Calculate:Ultimate loadsDesign moment (M )Design shear force

Estimate effective depth andwidth of beam (see 3.9.1.5)

Calculate ultimate momentof resistance, Mu = 0.156f cubd 2

Beam is singlyreinforced(see 3.9.1.1)

YESNO IsMu > M

Beam is doublyreinforced(see 3.9.2)

Design shear reinforcement(see 3.9.1.3)

Check deflection(see 3.9.1.4)

Produce reinforcement details(see 3.9.1.6)

Fig. 3.48 Beam design procedure.

3.10.1 TYPES OF SLABSSlabs may be solid, ribbed, precast or in-situ and ifin-situ they may span two-ways. In practice, thechoice of slab for a particular structure will largelydepend upon economy, buildability, the loadingconditions and the length of the span. Thus forshort spans, generally less than 5 m, the most

economical solution is to provide a solid slab ofconstant thickness over the complete span (Fig.3.51).

With medium size spans from 5 to 9 m it ismore economical to provide flat slabs since they aregenerally easier to construct (Fig. 3.52). The easeof construction chiefly arises from the fact that the

9780415467193_C03a 9/3/09, 1:01 PM94

95

Slabs

Fig. 3.49 Floor slab as a series of beams connectedtransversely.

Fig. 3.51 Solid slab.

Flat roof

Laterally loadedwalls

Ground slabStairway

Column/wallfooting

Floor slab

Fig. 3.50 Various applications for slabs in reinforcedconcrete structures.

Fig. 3.52 Flat slab.

floor has a flat soffit. This avoids having to erectcomplicated shuttering, thereby making possiblespeedier and cheaper construction. The use of flatslab construction offers a number of other advan-tages, absent from other flooring systems, includingreduced storey heights, no restrictions on the posi-tioning of partitions, windows can extend up to theunderside of the slab and ease of installation ofhorizontal services. The main drawbacks with flatslabs are that they may deflect excessively and arevulnerable to punching failure. Excessive deflectioncan be avoided by deepening slabs or by thicken-ing the slab near the columns, using drop panels.

Punching failure arises from the fact that high liveloads results in high shear stresses at the supportswhich may allow the columns to punch throughthe slab unless appropriate steps are taken. Usingdeep slabs with large diameter columns, providingdrop panels and/or flaring column heads (Fig. 3.53),can avoid this problem. However, all these methodshave drawbacks, and research effort has thereforebeen directed at finding alternative solutions. Theuse of shear hoops, ACI shear stirrups, shearladders and stud rails (Fig. 3.54) are just a few ofthe solutions that have been proposed over recentyears. All are designed to overcome the problem offixing individual shear links, which is both labourintensive and a practical difficulty.

Shear hoops are prefabricated cages of shearreinforcement which are attached to the main steel.They are available in a range of diameters and aresuitable for use with internal and edge columns.Although superficially attractive, use of this systemhas declined significantly over recent years.

The use of ACI shear stirrups is potentiallythe simplest and cheapest method of preventingpunching shear in flat slabs. The shear stirrups arearrangements of conventional straight bars and linksthat form a ‘ ’, ‘T’ or ‘L’ shape for an internal,

9780415467193_C03a 9/3/09, 1:02 PM95


96

(a) (b) (c)

Fig. 3.53 Methods of reducing shear stresses in flat slab construction: (a) deep slab and large column; (b) slab with flaredcolumn head; (c) slab with drop panel and column head.

Direction of T1 slabreinforcement

Shear hoop type SS

� 0.5d

� 0.5d

Direction of T1slab reinforcement

Shear hoop type SS

High tensileribbed steel bars

Stud rail

Spacing bar

Shear ladder

Lacer bars

(c)

(b) (d)

(a)

Fig. 3.54 Prefabricated punching shear reinforcement for flat slabs: (a) shear hoops (b) ACI shear stirrups (c) shearladders (d) stud rails. Typical arrangements for an internal column.

Shear ladders are rows of traditional links thatare welded to lacer bars. The links resist the shearstresses and the lacer bars anchor the links tothe main steel. Whilst they are simple to design

edge or corner column respectively. The stirrupswork in exactly the same way as conventional shearreinforcement but can simply be attached to themain steel via the straight bars.

9780415467193_C03a 9/3/09, 1:02 PM96

97

Slabs

Fig. 3.57 One-way spanning solid slab: (a) plan; (b) elevation.

(a) (b)

Fig. 3.55 Ribbed slab.

and use they can cause problems of congestion ofreinforcement.

Stud rails are prefabricated high tensile ribbedheaded studs, which are held at standard centresby a welded spacer bar. These rails are arranged ina radial pattern and held in position during theconcrete pour by tying to either the top or bottomreinforcement. The studs work through directmechanical anchorage provided by their heads.They are easy to install but quite expensive.

With medium to long spans and light to moderatelive loads from 3 to 5 kN/m2, it is more economicalto provide ribbed slabs constructed using glassreinforced polyester, polypropylene or encapsu-lated expanded polystyrene moulds (Fig. 3.55).Such slabs have reduced self-weight compared tosolid slabs since part of the concrete in the tensionzone is omitted. However, ribbed slabs have higherformwork costs than the other slabs systems men-tioned above and, generally, they are found to beeconomic in the range 8 to 12 m.

With the emphasis on speed of erection andeconomy of construction, the use of precast con-crete floor slabs is now also popular with bothclients and designers. Fig. 3.56 shows two types ofprecast concrete units that can be used to formfloors. The hollow core planks are very commonas they are economic over short, medium and longspans. If desired the soffit can be left exposedwhereas the top is normally finished with a level-ling screed or appropriate flooring system. Cranageof large precast units, particularly in congested citycentre developments, is the biggest obstacle to thistype of floor construction.

The span ranges quoted above generally assumethe slab is supported along two opposite edges, i.e.it is one-way spanning (Fig. 3.57). Where longer

Fig. 3.56 Precast concrete floor units: (a) hollow coreplank (b) double ‘T’ unit.

(a)

(b)

in-situ concrete floor spans are required it is usu-ally more economical to support the slab on allfour sides. The cost of supporting beams or wallsneeds to be considered though. Such slabs arereferred to as two-way spanning and are normallydesigned as two-dimensional plates provided theratio of the length of the longer side to the lengthof the shorter side is equal to 2 or less (Fig. 3.58).

This book only considers the design of one-wayand two-way spanning solid slabs supporting uni-formly distributed loads. The reader is referred tomore specialised books on this subject for guid-ance on the design of the other slab types describedabove.

3.10.2 DESIGN OF ONE-WAY SPANNING SOLIDSLAB

The general procedure to be adopted for slabdesign is as follows:

1. Determine a suitable depth of slab.2. Calculate main and secondary reinforcement

areas.3. Check critical shear stresses.4. Check detailing requirements.

3.10.2.1 Depth of slab (clause 3.5.7, BS 8110)Solid slabs are designed as if they consist of aseries of beams of l metre width.

The effective span of the slab is taken as thesmaller of

9780415467193_C03a 9/3/09, 1:02 PM97


98

Wal

l B

Load

on

wal

l B

Load onwall A

Load onwall C

Wall A

�y

Load

on

wal

l D

Wal

l D �x

Wall C

Fig. 3.58 Plan of two-way spanning slab. lx length of shorter side, ly length of longer side. Provided ly/lx ≤ 2 slab will spanin two directions as indicated.

(a) the distance between centres of bearings, A,or

(b) the clear distance between supports, D, plusthe effective depth, d, of the slab (Fig. 3.59).

The deflection requirements for slabs, which arethe same as those for beams, will often control thedepth of slab needed. The minimum effective depthof slab, dmin, can be calculated using

dmin

spanbasic ratio modification factor

=×

(3.20)

The basic (span/effective depth) ratios are givenin Table 3.14. The modification factor is a functionof the amount of reinforcement in the slab which isitself a function of the effective depth of the slab.Therefore, in order to make a first estimate of theeffective depth, dmin, of the slab, a value of (say)1.4 is assumed for the modification factor. Themain steel areas can then be calculated (section3.10.2.2), and used to determine the actual valueof the modification factor. If the assumed valueis slightly greater than the actual value, the depthof the slab will satisfy the deflection requirementsin BS 8110. Otherwise, the calculation must berepeated using a revised value of the modificationfactor.

3.10.2.2 Steel areas (clause 3.5.4, BS 8110)The overall depth of slab, h, is determined byadding allowances for cover (Table 3.6) and halfthe (assumed) main steel bar diameter to the effec-tive depth. The self-weight of the slab together withthe dead and live loads are used to calculate thedesign moment, M.

The ultimate moment of resistance of the slab,Mu, is calculated using equation 3.11, developedin section 3.9.1.1, namely

Mu = 0.156fcubd 2

If Mu ≥ M, which is the usual conditionfor slabs, compression reinforcement will not berequired and the area of tensile reinforcement, As,is determined using equation 3.12 developed insection 3.9.1.1, namely

As =

Mf z0 y.87

where z = d[0.5 + ( . / . )0 25 0 9− K ] in which K =M/fcubd2.

Secondary or distribution steel is required in thetransverse direction and this is usually based onthe minimum percentages of reinforcement (As min)given in Table 3.25 of BS 8110:

As min = 0.24% Ac when fy = 250 N/mm2

As min = 0.13% Ac when fy = 500 N/mm2

where Ac is the total area of concrete.

3.10.2.3 Shear (clause 3.5.5 of BS 8110)Shear resistance is generally not a problem in solidslabs subject to uniformly distributed loads and,in any case, shear reinforcement should not be pro-vided in slabs less than 200 mm deep.

Fig. 3.59 Effective span of simply supported slab.

D

A

d

9780415467193_C03a 9/3/09, 1:02 PM98

99

As discussed for beams in section 3.9.1.3, thedesign shear stress, υ, is calculated from

v

Vbd

=

The ultimate shear resistance, υc, is determinedusing Table 3.11. If υ < υc, no shear reinforce-ment is required. Where υ > υc, the form and areaof shear reinforcement in solid slabs should beprovided in accordance with the requirementscontained in Table 3.21.

3. Crack width (clause 3.12.11.2.7, BS 8110).Unless the actual crack widths have been checkedby direct calculation, the following rules willensure that crack widths will not generally exceed0.3 mm. This limiting crack width is based onconsiderations of appearance and durability.

(i) No further check is required on bar spacing ifeither:(a) fy = 250 N/mm2 and slab depth ≤ 250 mm,

or(b) fy = 500 N/mm2 and slab depth ≤ 200 mm,

or(c) the reinforcement percentage (100A s/bd )

< 0.3%.(ii) Where none of conditions (a), (b) or (c) apply

and the percentage of reinforcement in the slabexceed 1 per cent, then the maximum cleardistance between bars (smax) given in Table 3.28of BS 8110 should be used, namely:

smax ≤ 280 mm when fy = 250 N/mm2

smax ≤ 155 mm when fy = 500 N/mm2

4. Curtailment of reinforcement (clause3.12.10.3, BS 8110). Simplified rules for the cur-tailment of reinforcement are given in clause3.12.10.3 of BS 8110. These are shown diagram-matically in Fig. 3.60 for simply supported andcontinuous solid slabs.

Slabs

Table 3.18 Form and area of shearreinforcement in solid slabs (Table 3.16,BS 8110)

Values of υ (N/mm 2) Area of shear reinforcementto be provided

υ < vc None requiredυc < υ < (υc + 0.4) Minimum links in

areas where υ > υc

Asv ≥ 0.4bsv /0.87fyv

(υc + 0.4) < υ < 0.8 fcu Design linksor 5 N/mm2 Asv ≥ bsv(υ − υc)/0.87fyv

50%

0.15��45ø

40% 100% 40%

100%0.15�

0.3�

50%

0.1� 0.2�

0.1� 0.1��

100%40% 40%

(a)

(b)

Fig. 3.60 Simplified rules for curtailment of bars in slabs:(a) simply supported ends; (b) continuous slab (based onFig. 3.25, BS 8110).

3.10.2.4 Reinforcement details (clause 3.12,BS 8110)

For reasons of durability the code specifies limitsin respect of:

1. minimum percentage of reinforcement2. spacing of reinforcement3. maximum crack widths.

These are outlined below together with thesimplified rules for curtailment of reinforcement.

1. Reinforcement areas (clause 3.12.5, BS8110). The area of tension reinforcement, As,should not be less than the following limits:

As ≥ 0.24%Ac when fy = 250 N/mm2

As ≥ 0.13%Ac when fy = 500 N/mm2

where Ac is the total area of concrete.

2. Spacing of reinforcement (clause 3.12.11.2.7,BS 8110). The clear distance between tension bars,sb, should lie within the following limits: hagg + 5 mmor bar diameter ≤ sb ≤ 3d or 750 mm whichever isthe lesser where hagg is the maximum aggregate size.(See also below section on crack widths.)

9780415467193_C03a 9/3/09, 1:03 PM99


100

4250 mm

150 mm

Example 3.11 Design of a one-way spanning concrete floor (BS 8110)A reinforced concrete floor subject to an imposed load of 4 kNm−2 spans between brick walls as shown below. Designthe floor for exposure class XC1 assuming the following material strengths:

fcu = 35 Nmm−2

fy = 500 Nmm−2

Overall depth of slab (h) = d + Φ/2 + c

= 155 + 10/2 + 25 = 185 mm

LOADING

DeadSelf weight of slab (gk) = 0.185 × 24 kNm−3 = 4.44 kNm−2

ImposedTotal imposed load (qk) = 4 kNm−2

Ultimate loadFor 1 m width of slab total ultimate load, W, is

= (1.4gk + 1.6qk) width of slab × span

= (1.4 × 4.44 + 1.6 × 4)1 × 4.25 = 53.62 kN

Design moment

M

W

. . . = =

×=

b

8

53 62 4 25

828 5 kNm

h

c

d

φ

DEPTH OF SLAB AND MAIN STEEL AREA

Overall depth of slab, h

Minimum effective depth, dmin =

spanbasic ratio modification factor ×

=

×=

(

4250

20152

say)1.4mm

Hence, assume effective depth of slab (d) = 155 mm. Assume diameter of main steel (Φ) = 10 mm. From Table 3.6,cover to all steel (c) for exposure class XC1 = 25 mm.

9780415467193_C03a 9/3/09, 1:03 PM100

101

Ultimate momentMu = 0.156fcubd2

= 0.156 × 35 × 103 × 1552

= 131.2 × 106 = 131.2 kNm

Since M < Mu, no compression reinforcement is required.

Main steel

K

Mf bd

.

.= =

×× ×

=cu

2

6

3 2

28 5 1035 10 155

0 0339

z = d [ . ( . / . ) ]0 5 0 25 0 9+ − K

= 155 [ . ( . . / . ) ]0 5 0 25 0 0339 0 9+ −

= 155 × 0.96 ≤ 0.95d (= 147 mm)

Hence z = 147 mm.

A

Mf zs

y

2mm /m width of slab .

.

. = =

×× ×

=0 87

28 5 100 87 500 147

4466

For detailing purposes this area of steel has to be transposed into bars of a given diameter and spacing usingsteel area tables. Thus from Table 3.22, provide 10 mm diameter bars spaced at 150 mm, i.e. H10 at 150 centres(As = 523mm2/m).

Slabs


Actual modification factorThe actual value of the modification can now be calculated using equations 7 and 8 given in Table 3.16 (section3.9.1.4).

Design service stress, f

f A

Asy s,req

s,pro

= 5

8 ν(equation 8, Table 3.16)

=

× ××

= −

.

5 500 446

8 523266 5 Nmm 2

Table 3.22 Cross-sectional area per metre width for various bar spacing (mm2)

Bar size Spacing of bars(mm)

50 75 100 125 150 175 200 250 300

6 566 377 283 226 189 162 142 113 94.38 1010 671 503 402 335 287 252 201 168

10 1570 1050 785 628 523 449 393 314 26212 2260 1510 1130 905 754 646 566 452 37716 4020 2680 2010 1610 1340 1150 1010 804 67020 6280 4190 3140 2510 2090 1800 1570 1260 105025 9820 6550 4910 3930 3270 2810 2450 1960 164032 16100 10700 8040 6430 5360 4600 4020 3220 268040 25100 16800 12600 10100 8380 7180 6280 5030 4190

9780415467193_C03a 9/3/09, 1:04 PM101


102

RA

V

RB

V

W

4.25 m

Modification factor

= +−

+

≤ . ( )

. .0 55

477

120 0 92 0

2

fM

bd

s (equation 7, Table 3.16)

= +−

+×

×

= . ( . )

. .

.0 55477 266 5

120 0 928 5 1010 155

1 396

3 2

Hence,

New dmin =

425020 1 39 .×

= 153 mm < assumed d = 155 mm

Minimum area of reinforcement, A s min, is equal to

A s min = 0.13%bh = 0.13% × 103 × 185 = 241 mm2/m < A s

Therefore take d = 155 mm and provide H10 at 150 mm centres as main steel.

SECONDARY STEELBased on minimum steel area = 241 mm2/m. Hence from Table 3.22, provide H8 at 200 mm centres (A s = 252 mm2/m).


SHEAR REINFORCEMENT

Design shear stress, υSince slab is symmetrically loaded

RA = RB = W/2 = 26.8 kN


υ

.

. = =

××

= −V

bd

26 8 10

10 1550 17

3

3Nmm 2

Design concrete shear stress, υcAssuming that 50 per cent of main steel is curtailed at the supports, A s = 523/2 = 262 mm2/m

100 100 26210 155

0 1693

Abd

s

.=

××

=

d = 165

Secondarysteel

Mainsteel

9780415467193_C03a 9/3/09, 1:04 PM102

103

08

23-H08-200

01 01

01-H10-150

Alternatebarsreversed

01-H10-150

40 155

30

10150

30

CL

0808

01425 mm

01-H10-150 Alternate barsreversed

23-H08-200

H08-200

Slabs

Example 3.11 continuedFrom Table 3.11, design concrete shear stress for grade 25 concrete is 0.44 Nmm−2. Hence

υc = (35/25)1/3 × 0.44 = 0.52 Nmm−2

From Table 3.16 since υ < υc, no shear reinforcement is required.

REINFORCEMENT DETAILSThe sketch below shows the main reinforcement requirements for the slab. For reasons of buildability the actualreinforcement details may well be slightly different.

Check spacing between barsMaximum spacing between bars should not exceed the lesser of 3d (= 465 mm) or 750 mm. Actual spacing = 150 mmmain steel and 200 mm secondary steel. OK

Maximum crack widthSince the slab depth does not exceed 200 mm, the above spacing between bars will automatically ensure that themaximum permissible crack width of 0.3 mm will not be exceeded.

9780415467193_C03a 9/3/09, 1:04 PM103


104

Example 3.12 Analysis of a one-way spanning concrete floor (BS 8110)A concrete floor reinforced with 10 mm diameter mild steel bars (fy = 250 N/mm2) at 125 mm centres (A s = 628 mm2

per metre width of slab) between brick walls as shown in Fig. 3.61. Calculate the maximum uniformly distributedimposed load the floor can carry.

10 mm barat 125 c/c

115 mm

Cover = 25 mm

h = 150 mm

h = 150 mm

f cu = 30 N mm−2

ρc = 24 kN m−3

150 mm

3000 mm

Fig. 3.61

EFFECTIVE SPANEffective depth of slab, d, is

d = h − cover − Φ/2

= 150 − 25 − 10/2 = 120 mm

Effective span is the lesser of

(a) centre to centre distance between bearings = 3000 mm(b) clear distance between supports plus effective depth = 2850 + 120 = 2970 mm.

Hence effective span = 2970 mm.

MOMENT CAPACITY, M

Assume z = 0.95d = 0.95 × 120 = 114 mm

A s =

Mf z0 87. y

Hence

M = A s·0.87fyz = 628 × 0.87 × 250 × 114

= 15.5 × 106 Nmm = 15.5 kNm per metre width of slab

MAXIMUM UNIFORMLY DISTRIBUTED IMPOSED LOAD (qk)

Loading

Dead loadSelf weight of slab (gk) = 0.15 × 24 kNm−3 = 3.6 kNm−2

9780415467193_C03a 9/3/09, 1:05 PM104

105

Slabs

Example 3.12 continuedUltimate loadTotal ultimate load (W ) = (1.4gk + 1.6qk)span

= (1.4 × 3.6 + 1.6qk)2.970

Imposed load

Design moment (M ) =

Wb8

From above, M = 15.5 kNm = (5.04 + 1.6qk)

2 9708

2.

Rearranging gives

qk

2kNm=× −

= − . / . .

. .

15 5 8 2 970 5 04

1 65 6

2

Lever arm (z)Check that assumed value of z is correct, i.e. z = 0.95d.

K =

Mf bdcu

2

6

3 2

15 5 1030 10 120

0 0359 .

.=

×× ×

=

z = d [ . ( . / . ) ]0 5 0 25 0 9+ − K ≤ 0.95d

= d [ . ( . . / . )]0 5 0 25 0 0359 0 9+ − = 0.958d

Hence, assumed value of z is correct and the maximum uniformly distributed load that the floor can carry is5.6 kNm−2.

Panel Bay

Fig. 3.62 Definition of panels and bays (Fig. 3.7,BS 8110).

3.10.3 CONTINUOUS ONE-WAY SPANNINGSOLID SLAB DESIGN

The design of continuous one-way spanning slabsis similar to that outlined above for single-spanslabs. The main differences are that (a) severalloading arrangements may need to be consideredand (b) such slabs are not statically determinate.Methods such as moment distribution can beused to determine the design moments and shearforces in the slab as discussed in section 3.9.3.1.However, where the following conditions aremet, the moments and shear forces can be calcu-lated using the coefficients in Table 3.12 of BS8110, part of which is reproduced here as Table3.23.

1. There are three or more spans of approximatelyequal length.

2. The area of each bay exceeds 30 m2 (Fig. 3.62).

3. The ratio of the characteristic imposed load to thecharacteristic dead load does not exceed 1.25.

4. The characteristic imposed load does not exceed5 kN/m2 excluding partitions.

9780415467193_C03a 9/3/09, 1:05 PM105


106

Table 3.23 Ultimate bending moments and shear forces in one-way spanning slabs with simple endsupports (Table 3.12, BS 8110)

End support End span Penultimate support Interior span Interior support

Moment 0 0.086Fb −0.086Fb 0.063Fb −0.063FbShear 0.4F – 0.6F – 0.5F

F = 1.4Gk + 1.6Q k; b = effective span

Example 3.13 Continuous one-way spanning slab design (BS 8110)Design the continuous one-way spanning slab in Example 3.10 assuming the cover to the reinforcement is 25 mm(Fig. 3.63).

LoadingDead load, gk = self-weight of slab + finishes = 0.15 × 24 + 1.5 = 5.1 kNm−2

Imposed load, qk = 4 kNm−2

For a 1 m width of slab, total ultimate load, F = (1.4gk + 1.6qk)width of slab × span = (1.4 × 5.1 + 1.6 × 4)1 × 3.75= 50.8 kN

Design moments and shear forcesSince area of each bay (= 8.5 × 15 = 127.5 m2) > 30 m2, qk/gk (= 4/5.1 = 0.78) < 1.25 and qk < 5 kNm−2, thecoefficients in Table 3.23 can be used to calculate the design moments and shear forces in the slab.

Position Bending moments (kNm) Shear forces (kN)

Supports 1 & 5 0 0.4 × 50.8 = 20.3Near middle of spans 1/2 & 4/5 0.086 × 50.8 × 3.75 = 16.4Supports 2 & 4 −0.086 × 50.8 × 3.75 = −16.4 0.6 × 50.8 = 30.5Middle of spans 2/3 & 3/4 0.063 × 50.8 × 3.75 = 12Support 3 −0.063 × 50.8 × 3.75 = −12 0.5 × 50.8 = 25.4

150 mm

3.75 3.75 3.75 3.75

1 2 3 4 5

Fig. 3.63

Steel reinforcement

Middle of span 1/2 (and 4/5)Assume diameter of main steel, φ = 10 mmEffective depth, d = hs − (φ/2 + c) = 150 − (10/2 + 25) = 120 mm

K

Mf bd

.

.= =

×× ×

=cu

2

6

2

16 4 1035 1000 120

0 0325

9780415467193_C03a 11/3/09, 11:05 AM106

107

Slabs

z d K ( . ( . / . ) ) ( . ( . . / . ))= + − = + −0 5 0 25 0 9 120 0 5 0 25 0 0325 0 9 = 115.5 ≤ 0.95d = 114 mm

A

M

f zs

y

2mm .

.

. =

× ×=

×× ×

=0 87

16 4 10

0 87 500 114331

6

> A s,min = 0.13%bh = 195 mm2 OK

From Table 3.22, provide H10@200 mm centres (A s = 393 mm2/m) in the bottom of the slab.

Support 2 (and 4)M = −16.4 kNm. Therefore, provide H10@200 mm centres in the top of the slab.

Middle of span 2/3 (and 3/4)M = 12 kNm and z = 0.95d = 114 mm. Hence

A

Mf z

Asy

2s,minmm

.

.

=× ×

=×

× ×=

0 8712 10

0 87 500 114242

6

� OK

Provide H10@300 mm centres (A s = 262 mm2/m) in bottom face of slab.

Support 3Since M = −12 kNm provide H10@300 mm centres in top face of slab.

Support 1 (and 5)According to clause 3.12.10.3.2 of BS 8110, although simple supports may have been assumed at end supports foranalysis, negative moments may arise which could lead to cracking. Therefore an amount of reinforcement equal tohalf the area of bottom steel at mid-span but not less than the minimum percentage of steel recommended in Table3.25 of BS 8110 should be provided in the top of the slab. Furthermore, this reinforcement should be anchored at thesupport and extend not less than 0.15b or 45 times the bar size into the span.

From above, area of reinforcement at middle of span 1/2 is 330 mm2/m. From Table 3.25 of BS 8110, the minimumarea of steel reinforcement is 0.13%bh = 0.0013 × 1000 × 150 = 195 mm2/m. Hence provide H10 at 300 mm centres(A s = 262 mm2/m) in the top of the slab.

Distribution steelBased on the minimum area of reinforcement = 195 mm2/m. Hence, provide H10 at 350 centres (A s = 224 mm2/m).

Shear reinforcement

Support 2 (and 4)Design shear force, V = 30.5 kN

ν

.

. = =

××

= −V

bd

30 5 10

1000 1200 25

3

Nmm 2

100 100 3931000 120

0 33A

bds

.=××

=

νc =

3525

3 × 0.57 = 0.64 Nmm−2 > ν.

From Table 3.21, no shear reinforcement is required.

Support 3Design shear force, V = 25.4 kN


9780415467193_C03a 9/3/09, 1:06 PM107


108

H10-300

H10-300

565 1125 1125 1125 1125

H10-200 H10-300

H10-200 H10-3003.75 m 3.75 m

1 2 3

Distribution steel is H10-350 (A s = 224 mm2/m)


ν

.

. = =

××

= −V

bd

25 4 10

1000 1200 21

3

Nmm 2

100 100 2621000 120

0 22A

bds

.=××

=

νc =

3525

3

× 0.51 = 0.57 Nmm−2 > ν OK

No shear reinforcement is necessary.

Deflection

Actual

spaneffective depth

.= =3750120

31 25

Exterior spansSteel service stress, fs, is

f f

A

As y

s,req

s,pro

2Nmm= = × × = − . 5

8

5

8500

331

393263 2

ν

Modification factor

= +−

+

= +−

+×

×

= .

. .

.

. .

.0 55477

120 0 90 55

477 263 2

120 0 916 4 1010 120

1 42

2

6

3 2

fM

bd

s

From Table 3.14, basic span to effective depth ratio is 26. Hence

permissible

span

effective depth = basic ratio × mod. factor = 26 × 1.42 = 37 > 31.25 OK

Interior spansSteel service stress, fs, is

f f

A

As y

s,req

s,pro

2Nmm= = × × = − . 5

8

5

8500

242

262288 6

ν

Modification factor

= +−

+

= +−

+××

= .

. .

.

.

.0 55477

120 0 90 55

477 288 6

120 0 912 10

10 120

1 45

2

6

3 2

fM

bd

s

Hence permissible

span

effective depth = basic ratio × mod. factor = 26 × 1.45 = 37.7 > 31.25 OK

9780415467193_C03a 9/3/09, 1:07 PM108

109

Slabs

3.10.4 TWO-WAY SPANNING RESTRAINEDSOLID SLAB DESIGN

The design of two-way spanning restrained slabs(Fig. 3.64) supporting uniformly distributed loadsis generally similar to that outlined above for one-way spanning slabs. The extra complication arisesfrom the fact that it is rather difficult to determinethe design bending moments and shear forces inthese plate-like structures. Fortunately BS 8110contains tables of coefficients (βsx, βsy, βvx, βvy) thatmay assist in this task (Tables 3.24 and 3.25). Thus,the maximum design moments per unit width ofrectangular slabs of shorter side Bx and longer sideBy are given by

msx = βsxnB2x (3.21)

msy = βsynB2y (3.22)

wheremsx maximum design ultimate moments either

over supports or at mid-span on strips ofunit width and span Bx (Fig. 3.65)

msy maximum design ultimate moments eitherover supports or at mid-span on strips ofunit width and span By

n total design ultimate load per unit area =1.4gk + 1.6qk

Similarly, the design shear forces at supportsin the long span direction, υsy, and short spandirection, υsx, may be obtained from the followingexpressions

υsy = βvynBx (3.23)

υsx = βvxnBx (3.24)

Table 3.24 Bending moment coefficients, βsx and βsy, for restrained slabs (based on Table 3.14,BS 8110)

Type of panel and moments considered Short span coefficients, βsx Long spanValues of by /bx coefficients,

βsy, for all1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0 values of by /bx

Interior panelsNegative moment at continuous edge 0.031 0.037 0.042 0.046 0.050 0.053 0.059 0.063 0.032Positive moment at mid-span 0.024 0.028 0.032 0.035 0.037 0.040 0.044 0.048 0.024

One long edge discontinuousNegative moment at continuous edge 0.039 0.049 0.056 0.062 0.068 0.073 0.082 0.089 0.037Positive moment at mid-span 0.030 0.036 0.042 0.047 0.051 0.055 0.062 0.067 0.028

Two adjacent edges discontinuousNegative moment at continuous edge 0.047 0.056 0.063 0.069 0.074 0.078 0.087 0.093 0.045Positive moment at mid-span 0.036 0.042 0.047 0.051 0.055 0.059 0.065 0.070 0.034

Fig. 3.64 Bending of two-way spanning slabs.

Fig. 3.65 Location of moments.

� y

m sx

m sy

�x

These moments and shears are considered to actover the middle three quarters of the panel width.The remaining edge strips, of width equal to one-eight of the panel width, may be provided withminimum tension reinforcement. In some cases,where there is a significant difference in the sup-port moments calculated for adjacent panels, it maybe necessary to modify the mid-span moments inaccordance with the procedure given in BS 8110.

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110

Table 3.25 Shear force coefficients, βvx and βvy, for restrained slabs (based on Table 3.15, BS 8110)

Type of panel and location βvx for values of by/bx βvy

1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0

Four edges continuousContinuous edge 0.33 0.36 0.39 0.41 0.43 0.45 0.48 0.50 0.33

One long edge discontinuousContinuous edge 0.36 0.40 0.44 0.47 0.49 0.51 0.55 0.59 0.36Discontinuous edge 0.24 0.27 0.29 0.31 0.32 0.34 0.36 0.38 –

Two adjacent edges discontinuousContinuous edge 0.40 0.44 0.47 0.50 0.52 0.54 0.57 0.60 0.40Discontinuous edge 0.26 0.29 0.31 0.33 0.34 0.35 0.38 0.40 0.26

MID-SPAN MOMENTS

LoadingTotal dead load, gk = finishes etc. + self-weight of slab = 1.25 + 0.180 × 24 = 5.57 kNm−2

Imposed load, qk = 4 kNm−2

Design load, n = 1.4gk + 1.6qk = 1.4 × 5.57 + 1.6 × 4 = 14.2 kNm−2

Externaledge offloor

5 m

5

4

3

2

1

7 m 7 m

3.75 m

5 m

3.75 m

CBA

Fig. 3.66

Example 3.14 Design of a two-way spanning restrained slab (BS 8110)Fig. 3.66 shows a part plan of an office floor supported by monolithic concrete beams (not detailed), with individual slabpanels continuous over two or more supports. The floor is to be designed to support an imposed load of 4 kNm−2 andfinishes plus ceiling loads of 1.25 kNm−2. The characteristic strength of the concrete is 30 Nmm−2 and the steelreinforcement is 500 Nmm−2. The cover to steel reinforcement is 25 mm.

(a) Calculate the mid-span moments for panels AB2/3 and BC1/2 assuming the thickness of the floor is 180 mm.(b) Design the steel reinforcement for panel BC2/3 (shown hatched) and check the adequacy of the slab in terms of

shear resistance and deflection. Illustrate the reinforcement details on plan and elevation views of the panel.

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111

Slabs

PANEL AB2/3By inspection, panel AB2/3 has a discontinuous long edge. Also By/Bx = 7/5 = 1.4From Table 3.24,

short span coefficient for mid-span moment, βsx = 0.051long span coefficient for mid-span moment, βsy = 0.028

Hence mid-span moment in the short span, msx = βsxnB2x = 0.051 × 14.2 × 52 = 18.1 kNm and mid-span moment in

the long span, msy = βsynB2x = 0.028 × 14.2 × 52 = 9.9 kNm

PANEL BC1/2By inspection, panel BC1/2 has two adjacent discontinuous edges and By/Bx = 7/3.75 = 1.87. From Table 3.24,

short span coefficient for mid-span moment, βsx = 0.0675long span coefficient for mid-span moment, βsy = 0.034

Hence mid-span moment in the short span, msx = βsxnB2x = 0.0675 × 14.2 × 3.752 = 13.5 kNm and mid-span moment

in the long span, msy = βsynB2x = 0.034 × 14.2 × 3.752 = 6.8 kNm

PANEL BC2/3

Design momentBy inspection, panel BC2/3 is an interior panel. By/Bx = 7/5 = 1.4From Table 3.24,

short span coefficient for negative (i.e. hogging) moment at continuous edge, βsx,n = 0.05short span coefficient for positive (i.e. sagging) moment at mid-span, βsx,p = 0.037

long span coefficient for negative moment at continuous edge, βsy,n = 0.032 andlong span coefficient for positive moment at mid-span, βsy,p = 0.024

Hence negative moment at continuous edge in the short span,

msx,n = βsx,nnB2x = 0.05 × 14.2 × 52 = 17.8 kNm;

positive moment at mid-span in the short span,

msx,p = βsx,pnB2x = 0.037 × 14.2 × 52 = 13.1 kNm;

negative moment at continuous edge in the long span,

msy,n = βsy,nnB2x = 0.032 × 14.2 × 52 = 11.4 kNm;

and positive moment at mid-span in the long span,

msy,p = βsy,pnB2x = 0.024 × 14.2 × 52 = 8.5 kNm.

Steel reinforcement

Continuous supportsAt continuous supports the slab resists hogging moments in both the short-span and long-span directions. Thereforetwo layers of reinforcement will be needed in the top face of the slab. Comparison of design moments shows that themoment in the short span (17.8 kNm) is greater than the moment in the long span (11.4 kNm) and it is appropriatetherefore that the steel in the short span direction (i.e. main steel) be placed at a greater effective depth than thesteel in the long-span direction (i.e. secondary steel) as shown.


9780415467193_C03a 9/3/09, 1:09 PM111


112

Main steel(in short span)

Secondary steel(in long span)

d d ′

Assume diameter of main steel, φ = 10 mm and nominal cover, c = 25 mm. Hence,

Effective depth of main steel, d = h −

φ2

− c = 180 −

102

− 25 = 150 mm

Assume diameter of secondary steel, φ′ = 10 mm. Hence,

Effective depth of secondary steel, d ′ = h − φ −

φ′2

− c = 180 − 10 −

102

− 25 = 140 mm

Main steel

K

mf bd

.

.= =

×× ×

=sx,n

cu2

6

3 2

17 8 1030 10 150

0 0264

z d K ( . ( . / . ))= + −0 5 0 25 0 9 ≤ 0.95d = 0.95 × 150 = 142.5 mm

= + − ( . ( . . / . ))150 0 5 0 25 0 0264 0 9 = 150 × 0.97 = 146 mm

A

Mf zs

y

.

.

. ( . )= =

×× ×0 8717 8 10

0 87 500 0 95 150

6

= 287 mm2/m > 0.13%bh = 234 mm2/m OK

Provide H10@250 centres (A s = 314 mm2/m) in short span direction.

Secondary steel

K

m

f bd

.

.= =

×× ×

=sy,n

cu2

6

3 2

11 4 10

30 10 1400 0194

z d K ( . ( . / . ))= + −0 5 0 25 0 9 ≤ 0.95d = 0.95 × 140 = 133 mm

= + − ( . ( . . / . ))140 0 5 0 25 0 0194 0 9 = 140 × 0.98 = 137 mm

(Note that for slabs generally, z = 0.95d)

A

mf zs

sy,n

y

mm /m .

.

. ( . ) =

× ×=

×× × ×

=0 87

11 4 100 87 500 0 95 140

1976

2

≥ 0.13%bh = 234 mm2/m

Provide H10@300 centres (A s = 262 mm2/m) in long span direction.

Mid-spanAt mid-span the slab resists sagging moments in both the short-span and long-span directions, necessitating twolayers of reinforcement in the bottom face of the slab too. Comparison of mid-span moments shows that the momentin the short span (13.1 kNm) is greater than the moment in the long span (8.5 kNm) and it is again appropriatetherefore that the steel in the short span direction (main steel) be placed at a greater effective depth than the steelin the long span direction (secondary steel) as shown.

Secondary steel(in long span)

Main steel(in short span)

d d ′


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113

Slabs

Example 3.14 continuedAssume diameter of main steel, φ = 10 mm and nominal cover, c = 25 mm. Hence

Effective depth of main steel, d = h −

φ2

− c = 180 −

102

− 25 = 150 mm

Assuming diameter of secondary steel, φ′ = 10 mm. Hence

Effective depth of secondary steel, d ′ = h − φ −

φ′2

− c = 180 − 10 −

102

− 25 = 140 mm

Main steel

A

mf zs

sx,p

y

.

. . ( . )

= =×

× ×0 8713 1 10

0 87 500 0 95 150

6

= 211 mm2/m

≥ A s,min = 0.13%bh = 234 mm2/m

Provide H10@300 centres (A s = 262 mm2/m) in short span direction.

Secondary steel

A

mf zs

sy,p

y

2mm /m .

.

. ( . ) =

× ×=

×× × ×

=0 87

8 5 100 87 500 0 95 140

1476

≥ A s,min = 0.13%bh = 234 mm2/m

Provide H10@300 centres (A s = 262 mm2/m) in long span direction.

ShearFrom Table 3.25,

long span coefficient, βvy = 0.33 andshort span shear coefficient, βvx = 0.43

Design load on beams B2/3 and C2/3, vsy = βvynbx = 0.33 × 14.2 × 5 = 23.4 kNm−1

Design load on beams 2B/C and 3B/C, vsx = βvxnbx = 0.43 × 14.2 × 5 = 30.5 kNm−1 (critical)

Shear stress, ν

ν

.

. = =

××

= −sx 2Nmmbd

30 5 10

10 1500 20

3

3

100 100 31410 150

0 213

Abd

s

.=

××

=

From Table 3.12, νc = × .

3025

0 483

= 0.51 Nmm−2

Since νc > ν no shear reinforcement is required

DeflectionFor two-way spanning slabs, the deflection check is satisfied provided the span/effective depth ratio in the shorterspan does not exceed the appropriate value in Table 3.14 multiplied by the modification factor obtained via equations7 and 8 of Table 3.16

Actual

spaneffective depth

.= =5000150

33 3

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114

3

Section A–A

additional bars not shown on plan3

2

1

additional bars notshown on plan

2

1

4

Section B–B

B

17H10-3-300T2

B

B

A A

23H10-1-300B1

Bar markBar location

28H10-4-250T1

Bar spacing

Bar diameter

Bar type

Number of bars

17H10-2-300B2

Fig. 3.67

Example 3.14 continuedService stress, fs, is

f f

A

As y

s,req

s,pro

2Nmm= = × × = − 5

8

5

8500

211

262252

ν

Modification factor

= +−

+

= +−

+×

×

= .

. .

. .

.0 55477

120 0 90 55

477 252

120 0 913 1 1010 150

1 81

2

6

3 2

fmbd

s

sx,p

Permissible

spaneffective depth

= 26 × 1.81 = 47 > actual OK

Reinforcement detailsFig. 3.67 shows a sketch of the main reinforcement details for panel BC2/3. For reasons of buildability the actualreinforcement details may well be slightly different.

9780415467193_C03a 9/3/09, 1:10 PM114

115

Foundations

(a) (b)

Fig. 3.69 Foundation failures: (a) sliding failure;(b) overturning failure.

Fig. 3.71 Strip footings: (a) footing supporting columns; (b) footing supporting wall.

Elevation

Plan

A B C D

NNNN = = =

(a) (b)

foundations may bear directly on the ground or besupported on piles. The choice of foundation typewill largely depend upon (1) ground conditions (i.e.strength and type of soil) and (2) type of structure(i.e. layout and level of loading).

Pad footings are usually square or rectangularslabs and used to support a single column(Fig. 3.70). The pad may be constructed using massconcrete or reinforced concrete depending on therelative size of the loading. Detailed design of padfootings is discussed in section 3.11.2.1.

Continuous strip footings are used to supportloadbearing walls or under a line of closely spacedcolumns (Fig. 3.71). Strip footings are designed aspad footings in the transverse direction and in the

Fig. 3.70 Pad footing: (a) plan; (b) elevation.

(a) (b)

Load fromroof to column

Load fromfloor to column

Load fromfloor to column

Load fromcolumn tofoundation

Foundation loads resisted by ground

Fig. 3.68 Loading on foundations.

3.11 FoundationsFoundations are required primarily to carry the deadand imposed loads due to the structure’s floors,beams, walls, columns, etc. and transmit and dis-tribute the loads safely to the ground (Fig. 3.68).The purpose of distributing the load is to avoid thesafe bearing capacity of the soil being exceededotherwise excessive settlement of the structure mayoccur.

Foundation failure can produce catastrophiceffects on the overall stability of a structure so thatit may slide or even overturn (Fig. 3.69). Suchfailures are likely to have tremendous financial andsafety implications. It is essential, therefore, thatmuch attention is paid to the design of this elementof a structure.

3.11.1 FOUNDATION TYPESThere are many types of foundations which arecommonly used, namely strip, pad and raft. The

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116

Typical sections throughraft foundations

(b) (c)

(a)

Plan

Fig. 3.72 Raft foundations. Typical sections through raft foundation: (a) flat slab; (b) flat slab and downstand;(c) flat slab and upstand.

Hardstrata

Softstrata

Fig. 3.73 Piled foundations.

longitudinal direction as an inverted continuousbeam subject to the ground bearing pressure.

Where the ground conditions are relativelypoor, a raft foundation may be necessary in orderto distribute the loads from the walls and columnsover a large area. In its simplest form this mayconsist of a flat slab, possibly strengthened byupstand or downstand beams for the more heavilyloaded structures (Fig. 3.72).

Where the ground conditions are so poor thatit is not practical to use strip or pad footings butbetter quality soil is present at lower depths, theuse of pile foundations should be considered(Fig. 3.73).

The piles may be made of precast reinforcedconcrete, prestressed concrete or in-situ reinforcedconcrete. Loads are transmitted from the piles to

the surrounding strata by end bearing and/or fric-tion. End bearing piles derive most of their carry-ing capacity from the penetration resistance of thesoil at the toe of the pile, while friction piles relyon the adhesion or friction between the sides of thepile and the soil.

3.11.2 FOUNDATION DESIGNFoundation failure may arise as a result of (a) allow-able bearing capacity of the soil being exceeded,or (b) bending and/or shear failure of the base.The first condition allows the plan-area of the baseto be calculated, being equal to the design loaddivided by the bearing capacity of the soil, i.e.

Ground =

design loadplan area

< bearing

(3.25)pressure capacity of soil

Since the settlement of the structure occurs dur-ing its working life, the design loadings to be con-sidered when calculating the size of the base shouldbe taken as those for the serviceability limit state(i.e. 1.0Gk + 1.0Qk). The calculations to determinethe thickness of the base and the bending and shearreinforcement should, however, be based on ulti-mate loads (i.e. l.4Gk + 1.6Qk). The design of apad footing only will be considered here. The readeris referred to more specialised books on this sub-ject for the design of the other foundation typesdiscussed above. However, it should be borne inmind that in most cases the design process wouldbe similar to that for beams and slabs.

3.11.2.1 Pad footingThe general procedure to be adopted for thedesign of pad footings is as follows:

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117

Foundations

Example 3.15 Design of a pad footing (BS 8110)A 400 mm square column carries a dead load (G k) of 1050 kN and imposed load (Q k) of 300 kN. The safe bearingcapacity of the soil is 170 kNm−2. Design a square pad footing to resist the loads assuming the following materialstrengths:

fcu = 35 Nmm−2 fy = 500 Nmm−2

PLAN AREA OF BASE

Loading

Dead loadAssume a footing weight of 130 kN

Total dead load (Gk) = 1050 + 130 = 1180 kN

Serviceability load

Design axial load (N ) = 1.0Gk + 1.0Q k = 1.0 × 1180 + 1.0 × 300 = 1480 kN

Plan area

Plan area of base =

Nbearing capacity of soil

m2 . = =1480170

8 70

Hence provide a 3 m square base (plan area = 9 m2)

Axial loads: deadimposed = 300 kN

= 1050 kN

Fig. 3.75 Critical sections for shear. (Load on shaded areas to be used in design.)

Punching shearperimeter =columnperimeter+ 8 × 1.5d

1.5d

1.5d

1.0d

Face shear

Transverse shear

Fig. 3.74 Critical section for bending.

Load onshaded areato be usedin design

1. Calculate the plan area of the footing using ser-viceability loads.

2. Determine the reinforcement areas required forbending using ultimate loads (Fig. 3.74).

3. Check for punching, face and transverse shearfailures (Fig. 3.75).

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118

Self-weight of footingAssume the overall depth of footing (h) = 600 mm

Self weight of footing = area × h × density of concrete

= 9 × 0.6 × 24 = 129.6 kN < assumed (130 kN)

BENDING REINFORCEMENT

Design moment, MTotal ultimate load (W ) = 1.4G k + 1.6Q k

= 1.4 × 1050 + 1.6 × 300 = 1950 kN

Earth pressure (ps) =

W

plan area of base kNm 2 = = −1950

9217


1300 400

217 kN/m2

1300

Maximum design moment occurs at face of column (M ) = =

×

.psb2 2

2217 1 300

2

= 183 kNm/m width of slab

Ultimate moment

Effective depthBase to be cast against blinding, hence cover (c) to reinforcement = 50 mm (see Table 3.8). Assume 20 mm diameter(Φ) bars will be needed as bending reinforcement in both directions.

d

Cover

Φ

Hence, average effective depth of reinforcement, d, is

d = h − c − Φ = 600 − 50 − 20 = 530 mm

Ultimate moment

Mu = 0.156fcubd2 = 0.156 × 35 × 103 × 5302

= 1534 × 106 Nmm = 1534 kNm

Since Mu > M no compression reinforcement is required.

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119

Foundations

Example 3.15 continuedMain steel

K

Mf bd

.= =

×× ×

=cu

2

6

2

183 1035 1000 530

0 0186

z d K [ . ( . / . )]= + −0 5 0 25 0 9

= + − [ . ( . . / . )]d 0 5 0 25 0 0186 0 9

= 0.979d ≤ 0.95d = 0.95 × 530 = 504 mm

A

M

f zs

y

2mm /m .

. = =

×× ×

=0 87

183 10

0 87 500 504835

6

Minimum steel area is

0.13%bh = 780 mm2/m < A s OK

Hence from Table 3.22, provide H20 at 300 mm centres (A s = 1050 mm2/m) distributed uniformly across the full widthof the footing parallel to the x–x and y–y axis (see clause 3.11.3.2, BS 8110).

CRITICAL SHEAR STRESSES

Punching shear

Critical perimeter, pcrit, is

= column perimeter + 8 × 1.5d

= 4 × 400 + 8 × 1.5 × 530 = 7960 mm

Area within perimeter is

(400 + 3d)2 = (400 + 3 × 530)2 = 3.96 × 106 mm2

Ultimate punching force, V, is

V = load on shaded area = 217 × (9 − 3.96) = 1094 kN

Design punching shear stress, υ, is

υ

. = =

××

= −V

p dcrit

2Nmm1094 10

7960 5300 26

3

100 100 105010 530

0 1983

Abd

s

.=

××

=

Hence from Table 3.11, design concrete shear stress, υc, is

υc = (35/25)1/3 × 0.37 = 0.41 Nmm−2

Since υc > υ, punching failure is unlikely and a 600 mm depth of slab is acceptable.

1.5d Criticalperimeter

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120

01-11H20-300alternate barsreversed

A

01-11H20-300alternate barsreversed

Column starterbars (not designed)

01

A

01

75 k

icke

r

01 0101 01

Section A–A

770 mm 530 mm

d


Ultimate shear force (V ) = load on shaded area = ps × area = 217(3 × 0.770) = 501 kNDesign shear stress, υ, is

υ υ

. = =

×× ×

= −V

bd

501 10

3 10 5300 32

3

3Nmm 2

c�

Hence no shear reinforcement is required.

REINFORCEMENT DETAILSThe sketch below shows the main reinforcement requirements for the pad footing.

Face shearMaximum shear stress (υmax) occurs at face of column. Hence

υmax

=×

=×

× ×

( )

Wcolumn perimeter d

1950 104 400 530

3

= 2.3 Nmm−2 < permissible (= 0.8 35 = 4.73 Nmm−2)

Transverse shear

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121

Retaining walls

Fig. 3.77 Gravity retaining walls: (a) mass concrete wall; (b) masonry wall.

Meshreinforcement

Weep holes

Porous pipe

Drainage layer

(a) (b)

Granularbackfill

Mass concretefooting

Fig. 3.76 Section through road embankment incorporating a retaining wall.

Existing ground level

Retaining wallnecessary toavoid demolitionof building

Natural groundslope

Building

3.12 Retaining wallsSometimes it is necessary to maintain a differencein ground levels between adjacent areas of land.Typical examples of this include road and railwayembankments, reservoirs and ramps. A commonsolution to this problem is to build a natural slopebetween the two levels. However, this is not alwayspossible because slopes are very demanding ofspace. An alternative solution which allows animmediate change in ground levels to be effected isto build a vertical wall which is capable of resistingthe pressure of the retained material. These struc-tures are commonly referred to as retaining walls(Fig. 3.76). Retaining walls are important elementsin many building and civil engineering projects andthe purpose of the following sections is to brieflydescribe the various types of retaining walls avail-able and outline the design procedure associatedwith one common type, namely cantilever retainingwalls.

3.12.1 TYPES OF RETAINING WALLSRetaining walls are designed on the basis that theyare capable of withstanding all horizontal pressuresand forces without undue movement arising fromdeflection, sliding or overturning. There are two

main categories of concrete retaining walls (a) grav-ity walls and (b) flexible walls.

3.12.1.1 Gravity wallsWhere walls up to 2 m in height are required, it isgenerally economical to choose a gravity retainingwall. Such walls are usually constructed of massconcrete with mesh reinforcement in the faces toreduce thermal and shrinkage cracking. Other con-struction materials for gravity walls include masonryand stone (Fig. 3.77).

Gravity walls are designed so that the resultantforce on the wall due to the dead weight and theearth pressures is kept within the middle thirdof the base. A rough guide is that the width ofbase should be about a third of the height of theretained material. It is usual to include a granularlayer behind the wall and weep holes near the baseto minimise hydrostatic pressure behind the wall.Gravity walls rely on their dead weight for strengthand stability. The main advantages with this typeof wall are simplicity of construction and ease ofmaintenance.

3.12.1.2 Flexible wallsThese retaining walls may be of two basic types,namely (i) cantilever and (ii) counterfort.

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122

Passivepressure Pp

Fp Surfacescast againstundisturbed ground pa

FA

z

Active pressure

3.12.2 DESIGN OF CANTILEVER WALLSGenerally, the design process involves ensuringthat the wall will not fail either due to foundationfailure or structural failure of the stem or base.Specifically, the design procedure involves the fol-lowing steps:

1. Calculate the soil pressures on the wall.2. Check the stability of the wall.3. Design the bending reinforcement.

As in the case of slabs, the design of retainingwalls is usually based on a 1 m width of section.

3.12.2.1 Soil pressuresThe method most commonly used for determiningthe soil pressures is based on Rankin’s formula,which may be considered to be conservative but isstraightforward to apply. The pressure on the wallresulting from the retained fill has a destabilisingeffect on the wall and is normally termed activepressure (Fig. 3.80). The earth in front of the wallresists the destabilising forces and is termed passivepressure.

The active pressure (pa) is given by

pa = ρkaz (3.26)

whereρ = unit weight of soil (kN/m3)ka = coefficient of active pressurez = height of retained fill.

Here ka is calculated using

ka

sin sin

=−+

11

φφ

(3.27)

where φ is the internal angle of friction of retainedsoil. Typical values of ρ and φ for various soil typesare shown in Table 3.26.

Counterforts

Fig. 3.79 Counterfort retaining wall.

(i) Cantilever walls. Cantilevered reinforced con-crete retaining walls are suitable for heights up toabout 7 m. They generally consist of a uniformvertical stem monolithic with a base slab (Fig. 3.78).A key is sometimes incorporated at the base of thewall in order to prevent sliding failure of the wall.The stability of these structures often relies on theweight of the structure and the weight of backfillon the base. This is perhaps the most commontype of wall and, therefore, the design of such wallsis considered in detail in section 3.12.2.

(ii) Counterfort walls. In cases where a higherstem is needed, it may be necessary to designthe wall as a counterfort (Fig. 3.79). Counterfortwalls can be designed as continuous slabs span-ning horizontally between vertical supports knownas counterforts. The counterforts are designed ascantilevers and will normally have a triangular ortrapezoidal shape. As with cantilever walls, stabil-ity is provided by the weight of the structure andearth on the base.

Fig. 3.80 Active and passive pressure acting on a wall.

Granularbackfill

Heel

KeyToe

Base

Stem

Weep holesand drain

Fig. 3.78 Cantilever wall.

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123

Soil heaving up

(a)

Surface cast againstundisturbed ground

FA

Soil heaving upSurfacecast againstundisturbedground

h1

FF

h2

FP

Slip circle

(c)(b)

Retaining walls

k

kpa

=+−

= sin sin

11

1φφ

(3.29)

3.12.2.2 StabilityFountain failure of the wall may arise due to (a)sliding or (b) rotation. Sliding failure will occur ifthe active pressure force (FA) exceeds the passivepressure force (FP) plus the friction force (FF) aris-ing at the base/ground interface (Fig. 3.81(a)) where

FA = 0.5pah1 (3.30)

FP = 0.5pph2 (3.31)

FF = µWt (3.32)

The factor of safety against this type of failureoccurring is normally taken to be at least 1.5:

F FF

F P

A

.

+≥ 1 5 (3.33)

Table 3.26 Values of ρ and φ

Material ρ (kN/m3) φ

Sandy gravel 17–22 35°–40°Loose sand 15–16 30°–35°Crushed rock 12–22 35°–40°Ashes 9–10 35°–40°Broken brick 15–16 35°–40°

Fig. 3.81 Modes of failure: (a) sliding; (b) overturning; (c) slip circle.

The passive pressure (pp) is given by

pp = ρkpz (3.28)

whereρ = unit weight of soilz = height of retained fillkp = coefficient of passive pressure and is

calculated using

9780415467193_C03a 9/3/09, 1:12 PM123


124

Rotational failure of the wall may arise due to:

1. the overturning effect of the active pressure force(Fig. 3.81(b));

2. bearing pressure of the soil being exceeded whichwill display similar characteristics to (1); or

3. failure of the soil mass surrounding the wall(Fig. 3.81(c)).

Failure of the soil mass (type (3)) will not beconsidered here but the reader is referred to anystandard book on soil mechanics for an explana-tion of the procedure to be followed to avoid suchfailures.

Failure type (1) can be checked by takingmoments about the toe of the foundation (A) asshown in Fig. 3.82 and ensuring that the ratio ofsum of restoring moments (∑Mres) and sum of over-turning moments (∑Mover) exceeds 2.0, i.e.

∑∑

MM

res

over

.≥ 2 0 (3.34)

Failure type (2) can be avoided by ensuring thatthe ground pressure does not exceed the allowablebearing pressure for the soil. The ground pressureunder the toe (ptoe) and the heel (pheel) of the basecan be calculated using

p

ND

MDtoe 2

= + 6

(3.35)

p

ND

MDheel 2

= − 6 (3.36)

provided that the load eccentricity lies within themiddle third of the base, that is

M/N ≤ D/6 (3.37)

whereM = moment about centre line of baseN = total vertical load (Wt)D = width of base

3.12.2.3 Reinforcement areasStructural failure of the wall may arise if the baseand stem are unable to resist the vertical and hor-izontal forces due to the retained soil. The area ofsteel reinforcement needed in the wall can be cal-culated by considering the ultimate limit states ofbending and shear. As was pointed out at the be-ginning of this chapter, cantilever retaining wallscan be regarded for design purposes as three can-tilever beams (Fig. 3.2) and thus the equationsdeveloped in section 3.9 can be used here.

The areas of main reinforcement (As) can becalculated using

A

Mf zsy0.87

=

whereM = design momentfy = reinforcement gradez = d ( . / . )0 25 0 9− K ]K = M/fcubd 2

The area of distribution steel is based on theminimum steel area (As) given in Table 3.25 of BS8110, i.e.

As = 0.13%Ac when fy = 500 Nmm−2

As = 0.24%Ac when fy = 250 Nmm−2

where Ac is the total cross-sectional area of concrete.

Wbx1 + Wwx2 + Wsx3

FAy1

� 2.0

y1 = h/3

FA

x3

x2

Ww Ws

Wbx1A

h

Fig. 3.82

9780415467193_C03a 9/3/09, 1:13 PM124

125

Retaining walls

Example 3.16 Design of a cantilever retaining wall (BS 8110)The cantilever retaining wall shown below is backfilled with granular material having a unit weight, ρ, of 19 kNm−3

and an internal angle of friction, φ, of 30°. Assuming that the allowable bearing pressure of the soil is 120 kNm−2, thecoefficient of friction is 0.4 and the unit weight of reinforced concrete is 24 kNm−3

1. Determine the factors of safety against sliding and overturning.2. Calculate ground bearing pressures.3. Design the wall and base reinforcement assuming fcu = 35 kNm−2, fy = 500 kNm−2 and the cover to reinforcement

in the wall and base are, respectively, 35 mm and 50 mm.

5000

400A

700 400 2900

WW W s

Wb

FA

ka

1 − sin φ1 + sin φ

=

1 − sin 30°1 + sin 30°

=

1 − 0.5

1 + 0.5= =

1

3

Activepressure (pa) = kaρh

= 1/3 × 19 × 5.4= 34.2 kN m−2

SLIDINGConsider the forces acting on a 1 m length of wall. Horizontal force on wall due to backfill, FA, is

FA = 0.5pah = 0.5 × 34.2 × 5.4 = 92.34 kN

and

Weight of wall (Ww) = 0.4 × 5 × 24 = 48.0 kN

Weight of base (Wb) = 0.4 × 4 × 24 = 38.4 kN

Weight of soil (Ws) = 2.9 × 5 × 19 = 275.5 kN

Total vertical force (Wt) = 361.9 kN

Friction force, FF, is

FF = µWt = 0.4 × 361.9 = 144.76 kN

Assume passive pressure force (FP) = 0. Hence factor of safety against sliding is

144 7692 34

..

= 1.56 > 1.5 OK

OVERTURNINGTaking moments about point A (see above), sum of overturning moments (Mover) is

FA kNm×

=×

= .

. .

. 5 4

392 34 5 4

3166 2

9780415467193_C03b 9/3/09, 1:14 PM125

Design of reinforced concrete elements to BS 8110

126

Example 3.16 continuedSum of restoring moments (Mres) is

Mres = Ww × 0.9 + Wb × 2 + Ws × 2.55

= 48 × 0.9 + 38.4 × 2 + 275.5 × 2.55 = 822.5 kNm

Factor of safety against overturning is

822.5166.2

. .= 4 9 2 0� OK

GROUND BEARING PRESSUREMoment about centre line of base (M) is

M =

FA × .5 43

+ WW × 1.1 − WS × 0.55

=

×

. .92 34 5 43

+ 48 × 1.1 − 275.5 × 0.55 = 67.5 kNm

N = 361.9 kN

MN

mD

..

. . = = = =67 5361 9

0 1876

46

0 666 m�

Therefore, the maximum ground pressure occurs at the toe, ptoe, which is given by

ptoe = +

×

.

.361 94

6 67 542 = 116 kNm−2 < allowable (120 kNm−2)

Ground bearing pressure at the heel, pheel, is

pheel = −

×

.

.361 94

6 67 542 = 65 kNm−2

BENDING REINFORCEMENTWallHeight of stem of wall, hs = 5 m. Horizontal force on stem due to backfill, Fs, is

Fs = 0.5kaρh s2

= 0.5 × 1/3 × 19 × 52

= 79.17 kNm−1 width

Design moment at base of wall, M, is

M

F h

. . . = =

× ×=

γ f s s kNm3

1 4 79 17 53

184 7

Effective depthAssume diameter of main steel (Φ) = 20 mm.Hence effective depth, d, is

d = 400 − cover − Φ/2 = 400 − 35 − 20/2 = 355 mm

Ultimate moment of resistanceMu = 0.156fcubd 2 = 0.156 × 35 × 103 × 3552 × 10−6 = 688 kNm

Since Mu > M, no compression reinforcement is required.

9780415467193_C03b 9/3/09, 1:14 PM126

127

Retaining walls

Example 3.16 continuedSteel area

K

Mf bd

.

.= =

×× ×

=cu

2

6

3 2

184 7 1035 10 355

0 0419

z = d [ . ( . / . )]0 5 0 25 0 9+ − K

= 355 [ . ( . . / . )]0 5 0 25 0 0419 0 9+ − = 337 mm

A

M

f zs

y

2mm /m .

.

. = =

×× ×

=0 87

184 7 10

0 87 500 3371260

6

Hence from Table 3.22, provide H20 at 200 mm centres (A s = 1570 mm2/m) in near face (NF) of wall. Steel is alsorequired in the front face (FF) of wall in order to prevent excessive cracking. This is based on the minimum steelarea, i.e.

= 0.13%bh = 0.13% × 103 × 400 = 520 mm2/m

Hence, provide H12 at 200 centres (A s = 566 mm2)

Base

Heel

p3 = 91 +

2 9 162 4 914

. ( . )− = 142.8 kNm−2

Design moment at point C, Mc, is

385 7 2 92

2 9 38 4 1 4 1 454

91 2 92

51 8 2 9 2 92 3

160 52. .

. . . .

.

. . .

.

×+

× × ×−

×−

× ××

= kNm

Assuming diameter of main steel (Φ) = 20 mm and cover to reinforcement is 50 mm, effective depth, d, is

d = 400 − 50 − 20/2 = 340 mm

K

.

.=×

× ×=

160 5 1035 10 340

0 03976

3 2

z = 340 [ . ( . . / . )]0 5 0 25 0 0397 0 9+ − ≤ 0.95d = 323 mm

A

M

f zs

y

2mm /m .

.

. = =

×× ×

=0 87

160 5 10

0 87 500 3231142

6

Hence from Table 3.22, provide H20 at 200 mm centres (A s = 1570 mm2/m) in top face (T) of base.

p1 = 1.4 × 116 = 162.4 kN m−2

p2 = 1.4 × 65 = 91 kN m−2

HeelDToe

700 4002900

275.5 × 1.4 = 385.7 kN

CBA

p1 p3

9780415467193_C03b 9/3/09, 1:14 PM127


128

Distributionsteel H12-200

H12-200(FF)

200

kick

er

Starter bars H20-200H20-200 (T)

Distributionsteel H12-200 H12-200 (B)

H20-200 (NF)

(FF) far face(NF) near face(T) top face(B) bottom face

U-bars H12-200

ToeDesign moment at point B, MB, is given by

MB kNm≈

×−

× × ××

= . .

. . . .

.

162 4 0 72

0 7 38 4 1 4 0 74 2

36 52

A s

. .

=×36 5 1142

160 5 = 260 mm2/m < minimum steel area = 520 mm2/m

Hence provide H12 at 200 mm centres (A s = 566 mm2/m), in bottom face (B) of base and as distribution steel in baseand stem of wall.

REINFORCEMENT DETAILSThe sketch below shows the main reinforcement requirements for the retaining wall. For reasons of buildability theactual reinforcement details may well be slightly different.

Columns may be classified as short or slender,braced or unbraced, depending on various dimen-sional and structural factors which will be discussedbelow. However, due to limitations of space, thestudy will be restricted to the design of the mostcommon type of column found in building struc-tures, namely short-braced columns.

3.13.1 COLUMN SECTIONSSome common column cross-sections are shownin Fig. 3.84. Any section can be used, however,


3.13 Design of short bracedcolumns

The function of columns in a structure is to act asvertical supports to suspended members such asbeams and roofs and to transmit the loads fromthese members down to the foundations (Fig. 3.83).Columns are primarily compression membersalthough they may also have to resist bendingmoments transmitted by beams.

9780415467193_C03b 9/3/09, 1:14 PM128

129

Design of short braced columns

Loads incolumnstransmittedto foundation

Existinggroundlevel

Ground floor level

Foundation

First floor level

Loads from firstfloor slab/beamto column

Roof level

Load from roofto column

Fig. 3.83

procedures for the two column types are likely tobe different.

Clause 3.8.1.3 of BS 8110 classifies a column asbeing short if

lex

h < 15 and

ley

b < 15

whereBex effective height of the column in respect of

the major axis (i.e. x–x axis)Bey effective height of the column in respect of

the minor axisb width of the column cross-sectionh depth of the column cross-section

It should be noted that the above definition appliesonly to columns which are braced, rather thanunbraced. This distinction is discussed more fullyin section 3.13.3. Effective heights of columns iscovered in section 3.13.4.

3.13.3 BRACED AND UNBRACED COLUMNS(CLAUSE 3.8.1.5, BS 8110)

A column may be considered braced if the lateralloads, due to wind for example, are resisted by

provided that the greatest overall cross-sectionaldimension does not exceed four times its smallerdimension (i.e. h ≤ 4b, Fig. 3.84(c)). With sectionswhere h > 4b the member should be regarded as awall for design purposes (clause 1.2.4.1, BS 8110).

3.13.2 SHORT AND SLENDER COLUMNSColumns may fail due to one of three mechanisms:

1. compression failure of the concrete/steel rein-forcement (Fig. 3.85);

2. buckling (Fig. 3.86);3. combination of buckling and compression failure.

For any given cross-section, failure mode (1) ismost likely to occur with columns which are shortand stocky, while failure mode (2) is probable withcolumns which are long and slender. It is impor-tant, therefore, to be able to distinguish betweencolumns which are short and those which are slen-der since the failure mode and hence the design

(a) (b) (c)

b

h

Fig. 3.84 Column cross-sections.

Fig. 3.85

Fig. 3.86

9780415467193_C03b 9/3/09, 1:15 PM129


130

Fig. 3.90 Column end restraint conditions.

End condition 1

�0

Depth ofbeams �depth ofcolumn

As above

Base designed to resist moment

�0

�0

End condition 2

End condition 3

Depth of beams or slabs � depthof column

As above

Nominal restraint between beamsand column, e.g. beam designedand detailed as if simply supported

Base not designed to resist moment

Fig. 3.89 Columns unbraced in both directions.

between lateral restraints (bo) by a coefficient (β)which is a function of the fixity at the column endsand is obtained from Table 3.27.

be = βbo (3.38)

End condition 1 signifies that the column end isfully restrained. End condition 2 signifies that thecolumn end is partially restrained and end condi-tion 3 signifies that the column end is nominallyrestrained. In practice it is possible to infer thedegree of restraint at the column ends simply byreference to the diagrams shown in Fig. 3.90.

Table 3.27 Values of β for braced columns(Table 3.19, BS 8110)

End condition End condition at bottomof top

1 2 3

1 0.75 0.80 0.902 0.80 0.85 0.953 0.90 0.95 1.00

Shearwalls

Fig. 3.88 Columns braced in both directions.

y

x

Shear walls

Fig. 3.87 Columns braced in y direction and unbraced inthe x direction.

shear walls or some other form of bracing ratherthan by the column. For example, all the columnsin the reinforced concrete frame shown in Fig. 3.87are braced in the y direction. A column may beconsidered to be unbraced if the lateral loads areresisted by the sway action of the column. Forexample, all the columns shown in Fig. 3.87 areunbraced in the x direction.

Depending upon the layout of the structure, it ispossible for the columns to be braced or unbracedin both directions as shown in Figs 3.88 and 3.89respectively.

3.13.4 EFFECTIVE HEIGHTThe effective height (be) of a column in a givenplane is obtained by multiplying the clear height

9780415467193_C03b 11/3/09, 11:05 AM130

131


Example 3.17 Classification of a concrete columnDetermine if the column shown in Fig. 3.91 is short.

Fig. 3.91

�oy = 4.4 m

180 mm

�ox = 4.0 m

h = 350

x

y = 250

b

3.13.5 SHORT BRACED COLUMN DESIGNFor design purposes, BS 8110 divides short-bracedcolumns into three categories. These are:

1. columns resisting axial loads only;2. columns supporting an approximately symmet-

rical arrangement of beams;3. columns resisting axial loads and uniaxial or

biaxial bending.

Referring to the floor plan shown in Fig. 3.92, itcan be seen that column B2 supports beams whichare equal in length and symmetrically arranged.Provided the floor is uniformly loaded, columnB2 will resist an axial load only and is an exampleof category 1.

Column C2 supports a symmetrical arrangementof beams but which are unequal in length. ColumnC2 will, therefore, resist an axial load and moment.

However, provided that (a) the loadings on thebeams are uniformly distributed, and (b) the beamspans do not differ by more than 15 per cent of thelonger, the moment will be small. As such, columnC2 belongs to category 2 and it can safely be de-signed by considering the axial load only but usingslightly reduced values of the design stresses in theconcrete and steel reinforcement (section 3.13.5.2).

Columns belong to category 3 if conditions(a) and (b) are not satisfied. The moment herebecomes significant and the column may berequired to resist an axial load and uni-axial bend-ing, e.g. columns A2, B1, B3, C1, C3 and D2, oran axial loads and biaxial bending, e.g. A1, A3, D1and D3.

The design procedures associated with each ofthese categories are discussed in the subsectionbelow.

For bending in the y direction: end condition at top of column = 1, end condition at bottom of column = 1. Hencefrom Table 3.27, βx = 0.75.

b bex x ox

h h

. . = =

×=

β 0 75 4000350

8 57

For bending in the x direction: end condition at top of column = 2, end condition at bottom of column = 2. Hencefrom Table 3.27, βy = 0.85

b bey y oy

b b

. .= =

×=

β 0 85 4400250

14 96

Since both bex/h and bey/b are both less than 15, the column is short.

9780415467193_C03b 11/3/09, 11:06 AM131


132

Fig. 3.92 Floor plan.

3

2

1

A B C D�1 �1 �2

� 1� 1

N

Ac Asc

Fig. 3.93

3.13.5.1 Axially loaded columns(clause 3.8.4.3, BS 8110)

Consider a column having a net cross-sectionalarea of concrete Ac and a total area of longitudinalreinforcement Asc (Fig. 3.93).

As discussed in section 3.7, the design stresses forconcrete and steel in compression are 0.67fcu/1.5and fy/1.15 respectively, i.e.

Concrete design stress =

0 671 5.

.fcu = 0.45fcu

Reinforcement design stress =

fy

1.15 = 0.87fy

Both the concrete and reinforcement assist incarrying the load. Thus, the ultimate load N whichcan be supported by the column is the sum of theloads carried by the concrete (Fc) and the rein-forcement (Fs), i.e.

N = Fc + Fs

Fc = stress × area = 0.45fcuAc

Fs = stress × area = 0.87fy Asc

Hence, N = 0.45fcuAc + 0.87fy Asc (3.39)

Equation 3.35 assumes that the load is appliedperfectly axially to the column. However, in prac-tice, perfect conditions never exist. To allow for asmall eccentricity BS 8110 reduces the designstresses in equation 3.35 by about 10 per cent,giving the following expression:

N = 0.4fcuAc + 0.75fy Asc (3.40)

This is equation 38 in BS 8110 which can beused to design short-braced axially loaded columns.

3.13.5.2 Columns supporting an approximatelysymmetrical arrangement of beams(clause 3.8.4.4, BS 8110)

Where the column is subject to an axial load and‘small’ moment (section 3.13.5), the latter is takeninto account simply by decreasing the design stressesin equation 3.40 by around 10 per cent, giving thefollowing expression for the load carrying capacityof the column:

N = 0.35fcuAc + 0.67fy Asc (3.41)

This is equation 39 in BS 8110 and can be usedto design columns supporting an approximatelysymmetrical arrangement of beams provided (a) theloadings on the beams are uniformly distributed,and (b) the beam spans do not differ by more than15 per cent of the longer.

Equations 3.40 and 3.41 are not only used todetermine the load-carrying capacities of short-braced columns predominantly supporting axialloads but can also be used for initial sizing of theseelements, as illustrated in Example 3.18.

9780415467193_C03b 9/3/09, 1:15 PM132

133


Example 3.18 Sizing a concrete column (BS 8110)A short-braced column in which fcu = 30 Nmm−2 and fy = 500 Nmm−2 is required to support an ultimate axial load of2000 kN. Determine a suitable section for the column assuming that the area of longitudinal steel, A sc, is of the orderof 3 per cent of the gross cross-sectional area of column, A col.

Asc

2

Acol

Asc/2

Since the column is axially loaded use equation 3.40

N = 0.4fcuA c + 0.75fyA sc

2000 103

colcol col× = × −

+ × × . . 0 4 303100

0 75 5003100

AA A

A col = 87374 mm2

Assuming that the column is square,

b = h ≈ 87374 = 296 mm.

Hence a 300 mm square column constructed of concrete fcu = 30 Nmm−2 would be suitable.

Design charts are available for concrete grades 25,30, 35, 40, 45 and 50 and reinforcement grade460. For a specified concrete and steel strengththere is a series of charts for different d/h ratios inthe range 0.75 to 0.95 in 0.05 increments.

The construction of these charts can best beillustrated by considering how the axial load andmoment capacity of an existing column section isassessed. The solution to this problem is some-what simpler than normal column design as manyof the design parameters, e.g. grades of materialsand area and location of the reinforcement arepredefined. Nonetheless, both rely on an iterativemethod for solution. Determining the load capacityof an existing section involves investigating therelationship between the depth of neutral axisof the section and its axial load and co-existentmoment capacity. For a range of neutral axis depthsthe tensile and compressive forces acting on thesection are calculated. The size of these forcescan be evaluated using the assumptions previouslyoutlined in connection with the analysis of beamsections (3.9.1.1), namely:

3.13.5.3 Columns resisting axial load andbending

The area of longitudinal steel for columns resistingaxial loads and uniaxial or biaxial bending is nor-mally calculated using the design charts in Part 3of BS 8110. These charts are available for columnshaving a rectangular cross-section and a symmet-rical arrangement of reinforcement. BSI issued thesecharts when the preferred grade of reinforcementwas 460 not 500. Nevertheless, these charts couldstill be used to estimate the area of steel reinforce-ment required in columns but the steel areas ob-tained will be approximately 10 per cent greaterthan required. Fig. 3.94 presents a modified versionof chart 27 which takes account of the new gradeof steel reinforcement.

It should be noted that each chart is particularfor a selected

1. characteristic strength of concrete, fcu;2. characteristic strength of reinforcement, fy;3. d/h ratio.

9780415467193_C03b 9/3/09, 1:15 PM133


134

h = 400 mm

d = 350 mm

d ′ = 50 mm

x

X

2H32

b = 300 mm

2H32

X

16

N/b

h (N

mm

−2)

50

45

40

35

30

25

20

15

10

5

SeeExample 3.22

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 162 3 4 5 6 7 8 9 10 11 12 13 14 150

b

d h

Asc

2

fcu 30

fy 500

d/h 0.80

1

M/bh2 (N mm−2)

1Asc

2

0.4

1

2

34

5

6

7

8

100Asc /bh

Example 3.19 Analysis of a column section (BS 8110)Determine whether the column section shown in Fig. 3.95 is capable of supporting an axial load of 200 kN and amoment about the x–x axis of 200 kNm by calculating the load and moment capacity of the section when the depthof neutral axis of the section, x = ∞, 200 mm and 350 mm. Assume fcu = 35 Nmm−2 and fy = 500 Nmm−2.

Fig. 3.94 Column design chart (based on chart 27, BS 8110: Part 3).

4. The tensile strength of concrete is zero.

Once the magnitude and position of the tensileand compressive forces have been determined, theaxial load and moment capacity of the section canbe evaluated. Example 3.19 illustrates how theresults can be used to assess the suitability of thesection to resist a particular axial load and moment.

1. Sections that are plane before loading remainplane after loading.

2. The tensile and compressive stresses in the steelreinforcement are derived from Fig. 3.9.

3. The compressive stresses in concrete are basedon either the rectangular-parabolic stress blockfor concrete (Fig. 3.7) or the equivalent rectan-gular stress block (Fig. 3.16(e)).

Fig. 3.95

9780415467193_C03b 9/3/09, 1:15 PM134

135

x = 200 mm = h /2

Fsc

Fcc

Fst

0.9x

300 mm

X X

ε st

ε sc

d ′ = 50 mmε cu = 0.0035


Example 3.19 continuedLOAD AND MOMENT CAPACITY OF COLUMN WHEN X = ∞Assuming the simplified stress block for concrete, the stress and strain distributions in the section will be as shownin Fig. 3.96.

Fsc

Fcc

Fst

x = ∞

ε sc

ε st

XX

300 mm

50 mmε cu = 0.0035

Fig. 3.96

The compressive force in the concrete, Fcc, neglecting the area displaced by the reinforcement is

F

fbhcc

cu

mc

.

. .

=

=×

0 67 0 67 351 5γ × 300 × 400 = 1876 × 103 N

By inspection, ε sc = ε st = εcu = 0.0035 > ε y (= 0.0022, see Fig. 3.9). Hence

Fsc = Fst = 0.87fy(A sc/2) = 0.87 × 500 × (3216/2) = 699480 N

Axial load capacity of column, N, is

N = Fcc + Fsc + Fst = 1876 × 103 + 2 × 699480 = 3274960 N

The moment capacity of the section, M, is obtained by taking moments about the centre line of the section. Byinspection, it can be seen that M = 0.

LOAD AND MOMENT CAPACITY OF COLUMN WHEN X = 200 mmFig. 3.97 shows the stress and strain distributions when x = 200 mm.

Fig. 3.97

From similar triangles

ε εcu st

x d x

=

−

0 0035200 350 200.

=−εst ⇒ εst = 2.625 × 10−3 > ε y (= 0.0022)

9780415467193_C03b 9/3/09, 1:16 PM135


136

Example 3.19 continuedBy inspection εsc = εst. Hence the tensile force, Fst, and compressive force in the steel, Fsc, is

Fsc = Fst = 0.87fy(A sc/2) = 0.87 × 500 × (3216/2) = 699480 N

The compressive force in the concrete, Fcc, is

F

fxbcc

cu

mc=

=×

.

. .

.0 67

0 90 67 35

1 5γ × 0.9 × 200 × 300 = 844200 N


N = Fcc + Fsc − Fst = 844200 N

The moment capacity of the section, M, is again obtained by taking moments about the centre line of the section:

M = Fcc(h/2 − 0.9x/2) + Fst(d − h/2) + Fsc(h/2 − d ′)

= 844200(400/2 − 0.9 × 200/2) + 699480(350 − 400/2) + 699480(400/2 − 50) = 302.7 × 106 Nmm

LOAD AND MOMENT CAPACITY OF COLUMN WHEN X = 350 mmFig. 3.98 shows the stress and strain distributions when x = 350 mm.

x = 350 mm = d

Fsc

Fcc

Fst = 0

0.9x

b = 300 mm

X X

ε st = 0

ε sc

d ′ = 50 mmε cu = 0.0035

h /2

Fig. 3.98

As before

ε εcu st

x d x

=

−

0 0035350 350 350.

=−εst ⇒ εst = 0 and Fst = 0

Similarly, εsc = εcu (x − d ′)/x = 0.0035(350 − 50)/350 = 0.003 > ε y. Hence, the compressive force in the steel, Fsc, is

Fsc = 0.87fy(A sc/2) = 0.87 × 500 × (3216/2) = 699480 kN

F

fxbcc

cu

mc=

=×

.

. .

.0 67

0 90 67 35

1 5γ × 0.9 × 350 × 300 = 1477350 kN


N = Fcc + Fsc − Fst = 1477350 + 699480 − 0 = 2176830 N

The moment capacity of the section, M, is

M = Fcc(h/2 − 0.9x/2) + Fst(d − h/2) + Fsc(h/2 − d ′)

= 1477350(400/2 − 0.9 × 350/2) + 0 + 699480(400/2 − 50)

= 167.7 × 106 Nmm

9780415467193_C03b 9/3/09, 1:16 PM136

137

Example 3.19 continuedCHECK SUITABILITY OF PROPOSED SECTIONBy dividing the axial loads and moments calculated in (i)–(iii) by bh and bh2 respectively, the following values obtain:

x (mm) ∞ 200 350N/bh (Nmm−2) 27.3 7.0 18.1M/bh2 (Nmm−2) 0 6.3 3.5

Fig. 3.99 shows a plot of the results. By calculating N/bh and M/bh2 ratios for the design axial load (= 200 kN) andmoment (= 200 kNm) (respectively 16.7 and 4.2) and plotting on Fig. 3.99 the suitability of the section can bedetermined.

N/bh (N/mm2)

M/bh2

(N/mm2)

30

20

10

x = ∞ x = 350 mm

N/bh = 16.7M/bh2 = 4.2

x = 200 mm

1 2 3 4 5 6

Fig. 3.99

The results show that the column section is incapable of supporting the design loads. Readers may like to confirmthat if two 20 mm diameter bars (i.e. one in each face) were added to the section, the column would then havesufficient capacity.

Comparison of Figs 3.94 and 3.99 shows that the curves in both cases are similar and, indeed, if the area oflongitudinal steel in the section analysed in Example 3.19 was varied between 0.4 per cent and 8 per cent, a chart ofsimilar construction to that shown in Fig. 3.94 would result. The slight differences in the two charts arise from thefact that the d/h ratio and fcu are not the same.

Table 3.28 Values of the coefficient β(Table 3.22, BS 8110)

Nbhfcu

0 0.1 0.2 0.3 0.4 0.5 ≥0.6

β 1.00 0.88 0.77 0.65 0.53 0.42 0.30


(i) Uniaxial bending. With columns which aresubject to an axial load (N) and uni-axial moment(M), the procedure simply involves plotting theN/bh and M/bh2 ratios on the appropriate chartand reading off the corresponding area of rein-forcement as a percentage of the gross-sectionalarea of concrete (100Asc/bh) (Example 3.22). Wherethe actual d/h ratio for the section being designedlies between two charts, both charts may be readand the longitudinal steel area found by linearinterpolation.

(ii) Biaxial bending (clause 3.8.4.5, BS 8110).Where the column is subject to biaxial bending,the problem is reduced to one of uniaxial bendingsimply by increasing the moment about one of theaxes using the procedure outlined below. Referringto Fig. 3.100, if Mx/My ≥ h′/b′ the enhanced designmoment, about the x–x axis, Mx′, is

M ′x = Mx +

β ′′hb

My (3.42)

If Mx/My < h′/b′, the enhanced design moment aboutthe y-y axis, My′, is

M ′y = My +

β ′′b

hMx (3.43)

whereb′ and h′ are the effective depths (Fig. 3.100)β is the enhancement coefficient for biaxial

bending obtained from Table 3.28.

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138

(c) Spacing of reinforcement. The minimumdistance between adjacent bars should not be lessthan the diameter of the bars or hagg + 5 mm, wherehagg is the maximum size of the coarse aggregate.The code does not specify any limitation withregard to the maximum spacing of bars, butfor practical reasons it should not normally exceed250 mm.

3.13.6.2 Links (clause 3.12.7, BS 8110)The axial loading on the column may causebuckling of the longitudinal reinforcement andsubsequent cracking and spalling of the adjacentconcrete cover (Fig. 3.101). In order to preventsuch a situation from occurring, the longitudinalsteel is normally laterally restrained at regular inter-vals by links passing round the bars (Fig. 3.102).

Fig. 3.100

x

y

b

M x

x

y

My

b ′

h ′

h

Links necessaryto resist bucklingforces

Links

Fig. 3.102

Axial load maycause buckling oflongitudinalreinforcement

Fig. 3.101

The area of longitudinal steel can then bedetermined using the ultimate axial load (N) andenhanced moment (Mx′ or My′,) in the same way asthat described for uniaxial bending.

3.13.6 REINFORCEMENT DETAILSIn order to ensure structural stability, durabilityand practicability of construction BS 8110 laysdown various rules governing the minimum size,amount and spacing of (i) longitudinal reinforce-ment and (ii) links. These are discussed in thesubsections below.

3.13.6.1 Longitudinal reinforcement

(a) Size and minimum number of bars (clause3.12.5.4, BS 8110). Columns with rectangularcross-sections should be reinforced with a min-imum of four longitudinal bars; columns withcircular cross-sections should be reinforced with aminimum of six longitudinal bars. Each of the barsshould not be less than 12 mm in diameter.

(b) Reinforcement areas (clause 3.12.5, BS8110). The code recommends that for columnswith a gross cross-sectional area Acol, the area oflongitudinal reinforcement (Asc) should lie withinthe following limits:

0.4%Acol ≤ Asc ≤ 6%Acol in a vertically castcolumn and

0.4%Acol ≤ Asc ≤ 8%Acol in a horizontally castcolumn.

At laps the maximum area of longitudinal rein-forcement may be increased to 10 per cent of thegross cross-sectional area of the column for bothtypes of columns.

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139

H8-350 4H32

still widely observed in order to reduce the risk ofdiagonal shear failure of columns.

(b) Arrangement of links. The code furtherrequires that links should be so arranged thatevery corner and alternate bar in an outer layerof reinforcement is supported by a link passingaround the bar and having an included angle ofnot more than 135°. All other bars should be within150 mm of a restrained bar (Fig. 3.103).

(a) Size and spacing of links. Links should be atleast one-quarter of the size of the largest longitudi-nal bar or 6 mm, whichever is the greater. However,in practice 6 mm bars may not be freely availableand a minimum bar size of 8 mm is preferable.

Links should be provided at a maximum spacingof 12 times the size of the smallest longitudinalbar or the smallest cross-sectional dimension ofthe column. The latter condition is not mentionedin BS 8110 but was referred to in CP 114 and is

� 150 � 150 � 150 � 150

� 1

50�

150

� 1

50�

150

� 1

50

Fig. 3.103 Arrangement of links in columns.

Example 3.20 Axially loaded column (BS 8110)Design the longitudinal steel and links for a 350 mm square, short-braced column which supports the following axial loads:

G k = 1000 kN Qk = 1000 kN

Assume fcu = 40 Nmm−2 and fy & fyv = 500 Nmm−2.

LONGITUDINAL STEELSince column is axially loaded, use equation 3.40, i.e.

N = 0.4fcuAc + 0.75fyA sc

Total ultimate load (N ) = 1.4G k + 1.6Q k = 1.4 × 1000 + 1.6 × 1000 = 3000 kNSubstituting this into the above equation for N gives

3000 × 103 = 0.4 × 40 × (3502 − A sc) + 0.75 × 500A sc

A sc = 2897 mm2

Hence from Table 3.10, provide 4H32 (A sc = 3220 mm2)

LINKSThe diameter of the links is one-quarter times the diameter of the largest longitudinal bar, that is, 1/4 × 32 = 8 mm,but not less than 8mm diameter. The spacing of the links is the lesser of (a) 12 times the diameter of the smallestlongitudinal bar, that is, 12 × 32 = 384 mm, or (b) the smallest cross-sectional dimension of the column (= 350 mm).

Hence, provide H8 links at 350 mm centres.


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140

Example 3.21 Column supporting an approximately symmetricalarrangement of beams (BS 8110)

An internal column in a braced two-storey building supporting an approximately symmetrical arrangement of beams(350 mm wide × 600 mm deep) results in characteristic dead and imposed loads each of 1100 kN being applied to thecolumn. The column is 350 mm square and has a clear height of 4.5 m as shown in Fig. 3.104. Design the longitudinalreinforcement and links assuming

fcu = 40 Nmm−2 and fy & fyv = 500 Nmm−2

�0 = 4.5 m

600 mm

600 mm

Column 350 mm sq.

Fig. 3.104

CHECK IF COLUMN IS SHORT

Effective heightDepth of beams (600 mm) > depth of column (350 mm), therefore end condition at top of column = 1. Assumingthat the pad footing is not designed to resist any moment, end condition at bottom of column = 3. Therefore, fromTable 3.27, β = 0.9.

bex = bey = βbo = 0.9 × 4500 = 4050 mm

Short or slender

b bex ey

h b .= = =

4050350

11 6

Since both ratios are less than 15, the column is short.

LONGITUDINAL STEELSince column supports an approximately symmetrical arrangement of beams use equation 3.41, i.e.

N = 0.35fcuAc + 0.67fyA sc

Total axial load, N, is

N = 1.4Gk + 1.6Q k

= 1.4 × 1100 + 1.6 × 1100 = 3300 kN

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141

H8-3004H25

4H3212

635

0

Cover = 25 mm

Example 3.21 continuedSubstituting this into the above equation for N

3300 × 103 = 0.35 × 40(3502 − A sc) + 0.67 × 500A sc

⇒ Asc = 4938 mm2

Hence from Table 3.10, provide 4H32 and 4H25

(A sc = 3220 + 1960 = 5180 mm2)

LINKSThe diameter of the links is one-quarter times the diameter of the largest longitudinal bar, that is 1/4 × 32 = 8 mm,but not less than 8mm diameter. The spacing of the links is the lesser of (a) 12 times the diameter of the smallestlongitudinal bar, that is, 12 × 25 = 300 mm, or (b) the smallest cross-sectional dimension of the column (= 350 mm).Provide H8 links at 300 mm centres.

Example 3.22 Columns resisting an axial load and bending (BS 8110)Design the longitudinal and shear reinforcement for a 275 mm square, short-braced column which supports either

(a) an ultimate axial load of 1280 kN and a moment of 62.5 kNm about the x–x axis or(b) an ultimate axial load of 1280 kN and bending moments of 35 kNm about the x–x axis and 25 kNm about the

y–y axis.

Assume fcu = 30 Nmm−2, fy = 500 Nmm−2 and cover to all reinforcement is 35 mm.

LOAD CASE (A)

Longitudinal steel

Mbh2

6

2

62 5 10275 275

3 .

=×

×=

Nbh

=

××

=1280 10275 275

173

Assume diameter of longitudinal bars (Φ) = 20 mm, diameter of links (Φ′) = 8 mm

d = h − cover − Φ′ − Φ/2 = 275 − 35 − 8 − 20/2 = 222 mm

d/h = 222/275 = 0.8

From Fig. 3.94, 100A sc/bh = 3, A sc = 3 × 275 × 275/100 = 2269 mm2

Provide 8H20 (A sc = 2510mm2, Table 3.10)

LinksThe diameter of the links is one-quarter times the diameter of the largest longitudinal bar, that is, 1/4 × 20 = 5 mm,but not less than 8 mm diameter. The spacing of the links is the lesser of (a) 12 times the diameter of the smallest


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142

2H25

H8-275

2H25

4H20

H8-240

4H20

Example 3.22 continuedlongitudinal bar, that is, 12 × 20 = 240 mm, or (b) the smallest cross-sectional dimension of the column (= 275 mm).Provide H8 links at 240 mm centres

LOAD CASE (B)

Longitudinal steelAssume diameter of longitudinal bars (Φ) = 25 mm, diameter of links (Φ′) = 8 mm

b ′ = h ′ = h − Φ/2 − Φ′ − cover

= 275 − 25/2 − 8 − 35 = 220 mm

Mx/My = 35/25 = 1.4 > h ′/b ′ = 1

Nbhfcu

.=

×× ×

=1280 10

275 275 300 56

3

Hence β = 0.35 (Table 3.28)Enhanced design moment about x–x axis, M x′ , is

′ = +M M

hb

Mx x y β ′

′

= 35 +

0 35 220220

. × × 25 = 43.8 kNm

′ =×

×=

Mbh

x

2

6

2

43 8 10275 275

2 1 .

.

Nbh

=

××

=1280 10275 275

173

d/h = 220/275 = 0.8

From Fig. 3.94, 100A sc/bh = 2.2, A sc = 2.2 × 275 × 275/100 = 1664 mm2

Provide 4H25 (A sc = 1960 mm2, Table 3.10)

LinksThe diameter of the links is one-quarter times the diameter of the largest longitudinal bar, that is, 1/4 × 25 ≈ 6 mm,but not less than 8 mm diameter. The spacing of the links is the lesser of (a) 12 times the diameter of the smallestlongitudinal bar, that is, 12 × 25 = 300 mm, or (b) the smallest cross-sectional dimension of the column (= 275 mm).Provide H8 links at 275 mm centres.

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143

d = 600 mm

b = 900 mm

hf = 200 mm

2H25

b w = 300 mm

3.14 SummaryThis chapter has considered the design of a num-ber of reinforced concrete elements to BS 8110:Structural use of concrete. The elements consideredeither resist bending or axial load and bending.The latter category includes columns subject todirect compression and a combination of compres-sion and uni-axial or bi-axial bending. Elementswhich fall into the first category include beams,

slabs, retaining walls and foundations. They aregenerally designed for the ultimate limit states ofbending and shear and checked for the serviceabil-ity limit states of deflection and cracking. The no-table exception to this is slabs where the deflectionrequirements will usually be used to determine thedepth of the member. Furthermore, where slabsresist point loads, e.g. pad foundations or in flatslab construction, checks on punching shear willalso be required.

Summary

qk = 10 kN m−1

gk = 20 kN m−1

7 m

Questions1. (a) Derive from first principles the

following equation for the ultimatemoment of resistance (Mu) of a singlyreinforced section in which fcu is thecharacteristic cube strength ofconcrete, b is the width of the beamand d the effective depth

Mu = 0.156fcubd2

List any assumptions.(b) Explain what you understand by the

term ‘under-reinforced’ and whyconcrete beams are normally designedin this way.

2. (a) Design the bending reinforcement fora rectangular concrete beam whosebreadth and effective depth are 400 mmand 650 mm respectively to resist anultimate bending moment of 700 kNm.The characteristic strengths of theconcrete and steel reinforcement maybe taken as 40 N/mm2 and 500 N/mm2.

(b) Calculate the increase in moment ofresistance of the beam in (a) assumingthat two 25 mm diameter high-yieldbars are introduced into thecompression face at an effective depthof 50 mm from the extremecompression face of the section.

3. (a) Explain the difference between M andMu.

(b) Design the bending and shearreinforcement for the beam in Fig. Q3using the following information

fcu = 30 N/mm2 fy = 500 N/mm2

fyv = 250 N/mm b = 300 mm

span/depth ratio = 12

Fig. Q3

4. (a) Discuss how shear failure can arise inreinforced concrete members and howsuch failures can be avoided.

(b) Describe the measures proposed inBS 8110 to achieve durable concretestructures.

5. (a) A simply supported T-beam of 7 mclear span carries uniformly distributeddead (including self-weight) andimposed loads of 10 kN/m and15 kN/m respectively. The beam isreinforced with two 25 mm diameterhigh-yield steel bars as shown below.(i) Discuss the factors that influence

the shear resistance of a reinforcedconcrete beam without shearreinforcement. Assuming that theT-beam is made from grade 30concrete, calculate the beam’sshear resistance.

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144

(ii) Design the shear reinforcementfor the beam. Assumefyv = 500 N/mm2.

(b) Assuming As,req = 939 mm2 check thebeam for deflection.

6. Redesign the slab in Example 3.11assuming that the characteristic strengthof the reinforcement is 250 N/mm2.Comment on your results.

7. (a) Explain the difference betweencolumns which are short and slenderand those which are braced andunbraced.

(b) Calculate the ultimate axial loadcapacity of a short-braced columnsupporting an approximatelysymmetrical arrangement of beamsassuming that it is 500 mm squareand is reinforced with eight20 mm diameter bars. Assume thatfcu = 40 N/mm2, fy = 500 N/mm2 andthe concrete cover is 25 mm. Designthe shear reinforcement for thecolumn.

8. (a) A braced column which is 300 mmsquare is restrained such that it has aneffective height of 4.5 m. Classify thecolumn as short or slender.

(b) The column supports:(1) an ultimate axial load of 500 kN

and a bending moment of200 kNm, or

(2) an ultimate axial load of 800 kNand bending moments of 75 kNmabout the x-axis and 50 kNmabout the y-axis.

Design the longitudinal steel for bothload cases by constructing suitabledesign charts assuming fcu = 40 N/mm2,fy = 500 N/mm2 and the covers to allreinforcement is 35 mm.

9. An internal column in a multi-storeybuilding supporting an approximatelysymmetrical arrangement of beams carriesan ultimate load of 2,000 kN. The storeyheight is 5.2 m and the effective heightfactor is 0.85, fcu = 35 N/mm2 andfy = 500 N/mm2.

Assuming that the column is square,short and braced, calculate:1. a suitable cross-section for the column;2. the area of the longitudinal

reinforcement;3. the size and spacing of the links.Sketch the reinforcement detail incross-section.

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Chapter 4

Design in structuralsteelwork to BS 5950

This chapter is concerned with the design of structuralsteelwork and composite elements to British Standard5950. The chapter describes with the aid of a numberof fully worked examples the design of the followingelements: beams and joists, struts and columns, com-posite slabs and beams, and bolted and welded connec-tions. The section on beams and joists covers the designof members that are fully laterally restrained as well asmembers subject to lateral torsional buckling. The sectionon columns includes the design of cased columns andcolumn base plates.

4.1 IntroductionSeveral codes of practice are currently in use in theUK for design in structural steelwork. For build-ings the older permissible stress code BS 449 hasnow been largely superseded by BS 5950, which isa limit state code introduced in 1985. For steelbridges the limit state code BS 5400: Part 3 isused. Since the primary aim of this chapter is togive guidance on the design of structural steelworkelements, this is best illustrated by considering thecontents of BS 5950.

BS 5950 is divided into the following nine parts:

Part 1: Code of practice for design – Rolled and weldedsections.

Part 2: Specification for materials, fabrication anderection – Rolled and welded sections.

Part 3: Design in composite construction – Section 3.1:Code of practice for design of simple and con-tinuous composite beams.

Part 4: Code of practice for design of composite slabswith profiled steel sheeting.

Part 5: Code of practice for design of cold formed thingauge sections.

Part 6: Code of practice for design of light gauge pro-filed steel sheeting.

Part 7: Specification for materials, fabrication anderection – Cold formed sections and sheeting.

Part 8: Code of practice for fire resistant design.Part 9: Code of practice for stressed skin design.

Part 1 covers most of the material required foreveryday design. Since the majority of this chapteris concerned with the contents of Part 1, it shouldbe assumed that all references to BS 5950 refer toPart 1 exclusively. Part 3: Section 3.1 and Part 4deal with, respectively, the design of compositebeams and composite floors and will be discussedbriefly in section 4.10. The remaining parts of thecode generally either relate to specification or tospecialist types of construction, which are not rel-evant to the present discussion and will not bementioned further.

4.2 Iron and steel4.2.1 MANUFACTUREIron has been produced for several thousands ofyears but it was not until the eighteenth centurythat it began to be used as a structural material.The first cast-iron bridge by Darby was built in1779 at Coalbrookdale in Shropshire. Fifty yearslater wrought iron chains were used in ThomasTelford’s Menai Straits suspension bridge. How-ever, it was not until 1898 that the first steel-framedbuilding was constructed. Nowadays most construc-tion is carried out using steel which combines thebest properties of cast and wrought iron.

Cast iron is basically remelted pig iron whichhas been cast into definite useful shapes. Charcoalwas used for smelting iron but in 1740 AbrahamDarby found a way of converting coal into cokewhich revolutionised the iron-making process. Alater development to this process was the use oflimestone to combine with the impurities in theore and coke and form a slag which could be runoff independently of the iron. Nevertheless, the pigiron contained many impurities which made thematerial brittle and weak in tension.

9780415467193_C04 9/3/09, 1:21 PM145

Design in structural steelwork to BS 5950

146

In 1784, a method known as ‘puddling’ was de-veloped which could be used to convert pig iron intoa tough and ductile metal known as wrought iron.Essentially this process removed many of the impur-ities present in pig iron, e.g. carbon, manganese,silicon and phosphorus, by oxidation. However, itwas difficult to remove the wrought iron from thefurnace without contaminating it with slag. Thusalthough wrought iron was tough and ductile, it wasrather soft and was eventually superseded by steel.

Bessemer discovered a way of making steel fromiron. This involved filling a large vessel, lined withcalcium silicate bricks, with molten pig iron whichwas then blown from the bottom to remove impur-ities. The vessel is termed a converter and the pro-cess is known as acid Bessemer. However, thisconverter was unable to remove the phosphorus fromthe molten iron which resulted in the metal beingweak and brittle and completely unmalleable. Laterin 1878 Gilchrist Thomas proposed an alternativelining for the converter in which dolomite was usedinstead of silica and which overcame the problemsassociated with the acid Bessemer. His process be-came known as basic Bessemer. The resulting metal,namely steel, was found to be superior to cast ironand wrought iron as it had the same high strengthin tension and compression and was ductile.

4.2.2 PROPERTIESIf a rod of steel is subjected to a tensile test(Fig. 4.1(a)), and the stress in the rod (load/cross

sectional area in N/mm2) is plotted against the strain(change in length/original length), as the load isapplied, a graph similar to that shown in Fig. 4.1(b)would be obtained. Note that the stress-strain curveis linear up to a certain value, known as the yieldpoint. Beyond this point the steel yields without anincrease in load, although there is significant ‘strainhardening’ as the bar continues to strain towardsfailure. This is the plastic range.

In the elastic range the bar will return to its orig-inal length if unloaded. However, once past the yieldpoint, in the plastic range, the bar will be perman-ently strained after unloading. Fig. 4.1(c) shows theidealised stress-strain curve for structured steelworkwhich is used in the design of steel members.

The slope of the stress-strain curve in the elasticrange is referred to as the modulus of elasticity orYoung’s modulus and is denoted by the letter E. Itindicates the stiffness of the material and is used tocalculate deflections under load. Structural steelhas a modulus of elasticity of 205 kN/mm2.

4.3 Structural steel andsteel sections

Structural steel is manufactured in three basicgrades: S275, S355 and S460. Grade S460 is thestrongest, but the lower strength grade S275 is themost commonly used in structural applications,for reasons that will later become apparent. In this

Str

ess

(Nm

m−2

)

460

355

275

E

1

0.1 0.2 0.3

Strain

Strain hardening

Yield pointFailure

LOAD

Steel rod

(b)

(a)

Stress =Load (N)

Strain =Change in length

Yield point

Plastic range

Str

ess

Strain

Elastic range

(c)

Original length

Cross-sectional area

Fig. 4.1 Stress–strain curves for structural steel: (a) schematic arrangements of test; ( b) actual stress-strain curain curvefrom experiment; (c) idealised stress-strain relationship.

9780415467193_C04 9/3/09, 1:21 PM146

147

t

Bb

y

T

xd x

t

y

T

xd x x x

y

yy

y

B

b

Universal BeamBeams (and columns)

Universal ColumnColumns (and beams)

Channel SectionSmaller beams

Equal AngleStruts and ties

Circular Hollow SectionColumns and space frames

Rectangular Hollow SectionBridge beams

D D

Fig. 4.2 Standard rolled steel sections.

classification system ‘S’ stands for structural andthe number indicates the yield strength of the ma-terial in N/mm2.

Figure 4.2 shows in end view a selection of sec-tions commonly used in steel design together withtypical applications. Depending on the size and thedemand for a particular shape, some sections maybe rolled into shape directly at a steel rolling mill,while others may be fabricated in a welding shopor on site using (usually) electric arc welding.

These sections are designed to achieve economyof material while maximising strength, particularlyin bending. Bending strength can be maximised byconcentrating metal at the extremities of the section,where it can sustain the tensile and compressivestresses associated with bending. The most com-monly used sections are still Universal Beams (UBs)and Universal Columns (UCs). While boxes andtubes have some popularity in specialist applica-tions, they tend to be expensive to make and aredifficult to maintain, particularly in small sizes.

The geometric properties of these steel sections,including the principal dimensions, area, secondmoment of area, radius of gyration and elastic andplastic section moduli have been tabulated in abooklet entitled Structural Sections to BS4: Part 1:1993 and BS EN10056: 1999 which is publishedby Corus Construction and Industrial. Appendix Bcontains extracts from these tables for UBs andUCs, and will be frequently referred to. The axesx–x and y–y shown in Fig. 4.2 and referred to inthe steel tables in Appendix B denote the strongand weak bending axes respectively. Note that thesymbols used to identify particular dimensions ofuniversal sections in the booklet are not consistentwith the notation used in BS 5950 and have there-fore been changed in Appendix B to conform withBS 5950.

This chapter we will concentrate on the designof UBs and UCs and their connections to suitvarious applications. Irrespective of the elementbeing designed, the designer will need an

Structural steel and steel sections

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148

understanding of the following aspects which arediscussed next.

(a) symbols(b) general principles and design methods(c) loadings(d) design strengths.

4.4 SymbolsFor the purpose of this chapter, the followingsymbols have been used. These have largely beentaken from BS 5950.

GEOMETRIC PROPERTIESA area of sectionAg gross sectional area of steel sectionB breadth of sectionb outstand of flangeD depth of sectiond depth of webIx, Iy second moment of area about the

major and minor axesJ torsion constant of sectionL length of spanr root radiusrx, ry radius of gyration of a member about

its major and minor axesSeff effective plastic modulusSx, Sy plastic modulus about the major and

minor axesT thickness of flanget thickness of webu buckling parameter of the sectionx torsional index of sectionZeff effective elastic modulusZx, Zy elastic modulus about major and

minor axes

BENDINGAv shear areab1 stiff bearing lengthE modulus of elasticityFt tensile forceFv shear forceL actual lengthLE effective lengthM design moment or large end momentMc moment capacityMb buckling resistance momentMx maximum major axis momentPbw bearing resistance of an unstiffened webPs bearing resistance of stiffener

Px buckling resistance of an unstiffenedweb

Pxs buckling resistance of stiffenerk = T + rmLT equivalent uniform moment factor for

lateral torsional bucklingPcs contact stressPv shear capacity of a sectionpc compressive strength of steelpb bending strength of steelpy design strength of steelv slenderness factor for beamβ ratio of smaller to larger end momentβw a ratio for lateral torsional bucklingγ f overall load factorγm material strength factorδ deflectionε constant = (275/py)

1/2

λ slenderness ratioλLT equivalent slenderness

COMPRESSION

Abe effective area of baseplateAg gross sectional area of steel sectionc largest perpendicular distance from the

edge of the effective portion of thebaseplate to the face of the columncross-section

Fc ultimate applied axial loadL actual lengthLE effective lengthMb buckling resistance momentMcx, Mcy moment capacity of section about the

major and minor axes in the absenceof axial load

Mex, Mey eccentric moment about the major andminor axes

Mx, My applied moment about the major andminor axes

MLT maximum major axis moment in thesegment between restraints againstlateral torsional buckling

m equivalent uniform moment factorPc compression resistance of columnPs squash load of columnPcs compression resistance of short strutPE Euler loadpc compressive strengthpyp design strength of the baseplatetp thickness of baseplateω pressure under the baseplateλ slenderness ratioλLT equivalent slenderness

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149

CONNECTIONSa effective throat size of weldαe effective net areaαg gross areaαn net area of plateAs effective area of boltAt tensile stress area of a boltdb diameter of boltDh diameter of bolt holee1 edge distancee2 end distancep pitcht thickness of partFs applied shear forceFt applied tension forcePbb bearing capacity of boltPbg bearing capacity of the parts connected

by preloaded boltsPbs bearing capacity of parts connected by

boltsPs shear capacity of a boltPsL slip resistance provided by a preloaded

boltPt tension capacity of a member or boltPo minimum shank tensionpbb bearing strength of a boltpbs bearing strength of connected partsps shear strength of a boltpt tension strength of a boltpw design strength of a fillet-welds leg length of a fillet-weldKe coefficient = 1.2 for S275 steelK s coefficient = 1.0 for clearance holesµ slip factor

COMPOSITESAcv mean cross-sectional area of concreteAsv cross-sectional area of steel

reinforcementαe modular ratioBe effective breadth of concrete flange

Dp depth of profiled metal deckingDs depth of concrete flangeδ deflectionFv shear forceQ strength of shear studsRc compression resistance of concrete

flangeR f tensile resistance of steel flangeR s tensile resistance of steel beams longitudinal spacing of studsν longitudinal shear stress per unit lengthyp depth of neutral axis

4.5 General principles anddesign methods

As stated at the outset, BS 5950 is based on limitstate philosophy. Table 1 of BS 5950, reproducedas Table 4.1, outlines typical limit states appropri-ate to steel structures.

As this book is principally about the design ofstructural elements we will be concentrating on theultimate limit state of strength (1), and the service-ability limit state of deflection (5). Stability (2) isan aspect of complete structures or sub-structuresthat will not be examined at this point, except tosay that structures must be robust enough not tooverturn or sway excessively under wind or othersideways loading. Fatigue (3) is generally takenaccount of by the provision of adequate safetyfactors to prevent occurrence of the high stressesassociated with fatigue. Brittle fracture (4) can beavoided by selecting the correct grade of steel forthe expected ambient conditions. Avoidance ofexcessive vibration (6) and oscillations (7), are as-pects of structural dynamics and are beyond thescope of this book. Corrosion can be a serious prob-lem for exposed steelwork, but correct preparationand painting of the steel will ensure maximumdurability (8) and minimum maintenance duringthe life of the structure. Alternatively, the use ofweather resistant steels should be considered.

General principles and design methods

Table 4.1 Limit states (Table 1, BS 5950)

Ultimate Serviceability

1 Strength (including general yielding, rupture, 5 Deflectionbuckling and forming a mechanism)

2 Stability against overturning and sway 6 Vibrationstability

3 Fracture due to fatigue 7 Wind induced oscillation4 Brittle fracture 8 Durability

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150

Although BS 5950 does not specifically mentionfire resistance, this is an important aspect that fun-damentally affects steel’s economic viability com-pared to its chief rival, concrete. Exposed structuralsteelwork does not perform well in a fire. The highconductivity of steel together with the thin sectionsused causes high temperatures to be quickly reachedin steel members, resulting in premature failure dueto softening at around 600°C. Structural steelworkhas to be insulated to provide adequate fire resis-tance in multi-storey structures. Insulation mayconsist of sprayed treatment, intumescent coatings,concrete encasement or boxing with plasterboard.All insulation treatments are expensive. However,where the steel member is encased in concrete itmay be possible to take structural advantage of theconcrete, thereby mitigating some of the additionalexpenditure incurred (Section 4.9.6 ). Guidanceon the design of fire protection for members insteel framed buildings can be found in Part 8 ofBS 5950.

For steel structures three principal methods ofdesign are identified in clause 2.1.2 of BS 5950:

1. Simple design. The structure is regarded as hav-ing pinned joints, and significant moments arenot developed at connections (Fig. 4.3(a) ). Thestructure is prevented from becoming a mecha-nism by appropriate bracing using shear wallsfor instance. This apparently conservative as-sumption is a very popular method of design.

2. Continuous design. The joints in the structureare assumed to be able to fully transfer the forcesand moments in the members which they attach(Fig. 4.3(b) ). Analysis of the structure may beby elastic or plastic methods, and will be morecomplex than simple design. However the in-creasing use of micro-computers has made this

method more viable. In theory a more economicdesign can be achieved by this method, butunless the joints are truly rigid the analysis willgive an upper bound (unsafe) solution.

3. Semi-continuous design. The joints in the structureare assumed to have some degree of strength andstiffness but not provide complete restraint as inthe case of continuous design. The actual strengthand stiffness of the joints should be determinedexperimentally. Guidance on the design of semi-continuous frames can be found in the follow-ing Steel Construction Institute publications:(i) Wind-moment Design of Unbraced Composite

Frames, SCI-P264, 2000.(ii) Design of Semi-continuous Braced Frames,

SCI-P183, 1997.

4.6 LoadingAs for structural design in other media, the de-signer needs to estimate the loading to which thestructure may be subject during its design life.

The characteristic dead and imposed loads canbe obtained from BS 6399: Parts 1 and 3. Windloads should be determined from BS 6399: Part 2or CP3: Chapter V: Part 2. In general, a character-istic load is expected to be exceeded in only 5% ofinstances, or for 5% of the time, but in the case ofwind loads it represents a gust expected only onceevery 50 years.

To obtain design loading at ultimate limit statefor strength and stability calculations the charac-teristic load is multiplied by a load factor obtainedfrom Table 2 of BS 5950, part of which is repro-duced as Table 4.2. Generally, dead load is multi-plied by 1.4 and imposed vertical (or live) load by1.6, except when the load case considers wind load

Fig. 4.3 (a) Simple and ( b) continuous design.

HingesDead and imposed load

Win

d lo

adin

g

Win

d lo

adin

g

Bracing No bracing assumed

(a) (b)

Fully rigid joints

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151

Table 4.3 Design strengths py (Table 9,BS 5950)

Steel grade Thickness, less than Design strength,or equal to (mm) py (N/mm2)

S275 16 27540 26563 25580 245

100 235150 225

S355 16 35540 34563 33580 325

100 315150 295

S460 16 46040 44063 43080 410

100 400

To obtain design loading at serviceability limitstate for calculation of deflections the most adverserealistic combination of unfactored characteristicimposed loads is usually used. In the case of windloads acting together with imposed loads, only 80%of the full specified values need to be considered.

4.7 Design strengthsIn BS 5950 no distinction is made between char-acteristic and design strength. In effect the materialsafety factor γm = 1.0. Structural steel used in theUK is specified by BS 5950: Part 2, and strengthsof the more commonly used steels are given inTable 9 of BS 5950, reproduced here as Table 4.3.As a result of the residual stresses locked into themetal during the rolling process, the thicker thematerial, the lower the design strength.

Having discussed these more general aspects re-lating to structural steelwork design, the followingsections will consider the detailed design of beamsand joists (section 4.8), struts and columns (section4.9), composite floors and beams (section 4.10)and some simple bolted and welded connections(section 4.11).

4.8 Design of steel beamsand joists

Structural design of steel beams and joists primarilyinvolves predicting the strength of the member.This requires the designer to imagine all the waysin which the member may fail during its designlife. It would be useful at this point, therefore, todiscuss some of the more common modes of fail-ure associated with beams and joists.

also, in which case, dead, imposed and wind loadsare all multiplied by 1.2.

Several loading cases may be specified to give a‘worst case’ envelope of forces and moments aroundthe structure. In the design of buildings withoutcranes, the following load combinations shouldnormally be considered (clause 2.4.1.2, BS 5950):

1. dead plus imposed2. dead plus wind3. dead, imposed plus wind.

Design of steel beams and joists

Table 4.2 Partial factors for loads (Table 2, BS 5950)

Type of load and load combinations Factor, γf

Dead load 1.4Dead load acting together with wind and imposed load 1.2Dead load whenever it counters the effects of other loads 1.0Dead load when restraining sliding, overturning or uplift 1.0Imposed load 1.6Imposed load acting together with wind load 1.2Wind load 1.4Storage tanks, including contents 1.4Storage tanks, empty, when restraining sliding, overturning or uplift 1.0

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152

Elastic

Beam remains straightwhen unloaded

σ

Elastic

Beam remains straightwhen unloaded

σ yield

Partially plastic

Beam remains slightlybent when unloaded

σ yield σ yield

Fully plastic

Plastic hinge formed

σ

Below yield

σ yield

At yield point

σ yield

Partially plastic

σ yield

Fully plastic

Stresses with increasing bending momentat centre span

Fig. 4.4 Bending failure of a bearn.

Fig. 4.5 Local flange buckling failure.

M MCompression

Tension

Flange buckling failure4.8.1 MODES OF FAILURE

4.8.1.1 BendingThe vertical loading gives rise to bending of thebeam. This results in longitudinal stresses beingset up in the beam. These stresses are tensile inone half of the beam and compressive in the other.As the bending moment increases, more and moreof the steel reaches its yield stress. Eventually, allthe steel yields in tension and/or compression acrossthe entire cross section of the beam. At this pointthe beam cross-section has become plastic and itfails by formation of a plastic hinge at the pointof maximum moment induced by the loading. Fig-ure 4.4 reviews this process. Chapter 2 summariseshow classical beam theory is derived from theseconsiderations.

4.8.1.2 Local bucklingDuring the bending process outlined above, if thecompression flange or the part of the web subjectto compression is too thin, the plate may actuallyfail by buckling or rippling, as shown in Fig. 4.5,before the full plastic moment is reached.

4.8.1.3 ShearDue to excessive shear forces, usually adjacent tosupports, the beam may fail in shear. The beamweb, which resists shear forces, may fail as shown inFig. 4.6(a), as steel yields in tension and compres-sion in the shaded zones. The formation of plastichinges in the flanges accompanies this process.

4.8.1.4 Shear bucklingDuring the shearing process described above, if theweb is too thin it will fail by buckling or rippling inthe shear zone, as shown in Fig. 4.6(b).

4.8.1.5 Web bearing and bucklingDue to high vertical stresses directly over a supportor under a concentrated load, the beam web mayactually crush, or buckle as a result of these stresses,as illustrated in Fig. 4.7.

4.8.1.6 Lateral torsional bucklingWhen the beam has a higher bending stiffness inthe vertical plane compared to the horizontal plane,the beam can twist sideways under the load. Thisis perhaps best visualised by loading a scale rule on

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153

(a)

Plastic hingesin flanges

V

Shearzone

V

TensionTensionCom

pres

sion

(b)

Folds or buckles

Fig. 4.7 Web buckling and web bearing failures.

Buckling

Crushing

Support Support Support

Buc

klin

g

Crushing

Root of cantilever

Normal load

Normal load

Destabilizing load

End of cantilever

Fig. 4.8 Lateral torsional buckling of cantilever.

its edge, as it is held as a cantilever – it will tend totwist and deflect sideways. This is illustrated inFig. 4.8. Where a beam is not prevented frommoving sideways, by a floor, for instance, or thebeam is not nominally torsionally restrained at sup-ports, it is necessary to check that it is laterallystable under load. Nominal torsional restraint maybe assumed to exist if web cleats, partial depthend plates or fin plates, for example, are present(Fig. 4.9).

4.8.1.7 DeflectionAlthough a beam cannot fail as a result of exces-sive deflection alone, it is necessary to ensure thatdeflections are not excessive under unfactoredimposed loading. Excessive deflections are those


Fig. 4.6 Shear and shear buckling failures: (a) shear failure; ( b) shear buckling.

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154

(a) (b) (c)

Fig. 4.9 Nominal torsional restraint at beam support supplied by (a) web cleats (b) end plate (c) fin plate.

resulting in severe cracking in finishes which wouldrender the building unserviceable.

4.8.2 SUMMARY OF DESIGN PROCESSThe design process for a beam can be summarisedas follows:

1. determination of design shear forces, Fv, andbending moments, M, at critical points on theelement (see Chapter 2);

2. selection of UB or UC;3. classification of section;4. check shear strength; if unsatisfactory return to

(2);5. check bending capacity; if unsatisfactory return

to (2);6. check deflection; if unsatisfactory return to (2);7. check web bearing and buckling at supports or

concentrated load; if unsatisfactory provide webstiffener or return to (2);

8. check lateral torsional buckling (section 4.8.11);if unsatisfactory return to (2) or provide lateraland torsional restraints;

9. summarise results.

4.8.3 INITIAL SECTION SELECTIONIt is perhaps most often the case in the design ofskeletal building structures, that bending is thecritical mode of failure, and so beam bending theorycan be used to make an initial selection of section.Readers should refer to Chapter 2 for more clarifica-tion on bending theory if necessary.

To avoid bending failure, it is necessary to en-sure that the design moment, M, does not exceedthe moment capacity of the section, Mc, i.e.

M < Mc (4.1)

Generally, the moment capacity for a steel sec-tion is given by

Mc = pyS (4.2)

wherepy is the assumed design strength of the steelS is the plastic modulus of the section

Combining the above equations gives an expres-sion for S:

S > M /py (4.3)

This can be used to select suitable universal beamsections from steel tables (Appendix B) with theplastic modulus of section S greater than the cal-culated value.

4.8.4 CLASSIFICATION OF SECTIONHaving selected a suitable section, or proposed asuitable section fabricated by welding, it must beclassified.

4.8.4.1 Strength classificationIn making the initial choice of section, a steelstrength will have been assumed. If grade S275steel is to be used, for example, it may have beenassumed that the strength is 275 N/mm2. Now byreferring to the flange thickness T from the steeltables, the design strength can be obtained fromTable 9 of BS 5950, reproduced as Table 4.3.

If the section is fabricated from welded plate,the strength of the web and flange may be takenseparately from Table 9 of BS 5950 as that for theweb thickness t and flange thickness T respectively.

4.8.4.2 Section classificationAs previously noted, the bending strength of thesection depends on how the section performs inbending. If the section is stocky, i.e. has thickflanges and web, it can sustain the formation of aplastic hinge. On the other hand, a slender section,i.e. with thin flanges and web, will fail by localbuckling before the yield stress can be reached.Four classes of section are identified in clause 3.5.2of BS 5950:

Class 1 Plastic cross sections are those in which aplastic hinge can be developed with significantrotation capacity (Fig. 4.10). If the plastic designmethod is used in the structural analysis, all mem-bers must be of this type.

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155

Applied Moment

Mp

Me

Slender

Semi-compact

Compact

Plastic

Rotation

Class 2 Compact cross sections are those inwhich the full plastic moment capacity can bedeveloped, but local buckling may preventproduction of a plastic hinge with sufficient rota-tion capacity to permit plastic design.

Class 3 Semi-Compact cross sections can de-velop their elastic moment capacity, but localbuckling may prevent the production of the fullplastic moment.

Class 4 Slender cross sections contain slender ele-ments subject to compression due to moment oraxial load. Local buckling may prevent the fullelastic moment capacity from being developed.

Limiting width to thickness ratios for elements forthe above classes are given in Table 11 of BS 5950,part of which is reproduced as Table 4.4. (Refer toFig. 4.2 for details of UB and UC dimensions.)Once the section has been classified, the various

strength checks can be carried out to assess itssuitability as discussed below.

4.8.5 SHEARAccording to clause 4.2.3 of BS 5950, the shearforce, Fv, should not exceed the shear capacity ofthe section, Pv, i.e.

Fv ≤ Pv (4.5)

where Pv = 0.6py Av (4.6)

in which Av is the shear area (= tD for rolled I-,H- and channel sections). Equation (4.6) assumesthat the web carries the shear force alone.

Clause 4.2.3 also states that when the bucklingratio (d /t) of the web exceeds 70ε (see equation4.4), then the web should be additionally checkedfor shear buckling. However, no British universalbeam section, no matter what the grade, is affected.


Fig. 4.10 Typical moment/rotation characteristics of different classes of section.

Note. ε = (275/py)1/2 (4.4)

Table 4.4 Limiting width to thickness ratios (elements which exceed these limits are to be taken asclass 4, slender cross sections.) (based on Table 11, BS 5950)

Type of element (all rolled sections) Class of section

(1) Plastic (2) Compact (3) Semi-comp

Outstand element of compression flange

bT

≤ 9ε

bT

≤ 10ε

bT

≤ 15ε

Web with neutral axis at mid-depth

dt

≤ 80ε

dt

≤ 100ε

dt

≤ 120ε

Web where the whole cross-section isn/a n/a

dt

≤ 40εsubject to axial compression only

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156

4.8.6 LOW SHEAR AND MOMENT CAPACITYAs stated in clause 4.2.1.1 of BS 5950, at criticalpoints the combination of (i) maximum momentand co-existent shear and (ii) maximum shear andco-existent moment, should be checked.

If the co-existent shear force Fv is less than 0.6Pv,then this is a low shear load. Otherwise, if0.6Pv < Fv < Pv, then it is a high shear load.

When the shear load is low, the moment capa-city of the section is calculated according to clause4.2.5.2 of BS 5950 as follows:

For class 1 plastic or class 2 compact sections,the moment capacity

Mc = pyS ≤ 1.2pyZ (4.7)

wherepy design strength of the steelS plastic modulus of the sectionZ elastic modulus of the section

The additional check (Mc ≤ 1.2pyZ ) is to guardagainst plastic deformations under serviceabilityloads and is applicable to simply supported andcantilever beams. For other beam types this limitis 1.5pyZ.

For class 3 semi-compact sections

Mc = pyZ (4.8)

or alternatively Mc = pySeff ≤ 1.2pyZ (4.9)

where Seff is the effective plastic modulus (clause3.5.6 of BS 5950) and the other symbols are asdefined for equation 4.7.

Note that whereas equation 4.8 provides a con-servative estimate of the moment capacity of class 3compact sections, use of equation 4.9 is more effici-ent but requires additional computational effort.

For class 4 slender sections

Mc = pyZeff (4.10)

where Zeff is the effective elastic modulus (clause3.6.2 of BS 5950).

In practice the above considerations do not proveto be much of a problem. Nearly all sections ingrade S275 steel are plastic, and only a few sectionsin higher strength steel are semi-compact. No Brit-ish rolled universal beam sections in pure bending,no matter what the strength class, are slender orhave plastic or compact flanges and semi-compactwebs.

Example 4.1 Selection of a beam section in grade S275 steel(BS 5950)

The simply supported beam in Fig. 4.11 supports uniformly distributed characteristic dead and imposed loads of5 kN/m each, as well as a characteristic imposed point load of 30 kN at mid-span. Assuming the beam is fullylaterally restrained and there is nominal torsional restrain at supports, select a suitable UB section in S275 steel tosatisfy bending and shear considerations.

DESIGN BENDING MOMENT AND SHEAR FORCETotal loading = (30 × 1.6) + (5 × 1.4 + 5 × 1.6)10

= 48 + 15 × 10 = 198 kN

30 kN imposed load

5 kN/m dead load5 kN/m imposed load

A B

RA RB

C

10 metres

Fig. 4.11 Loading for example 4.1.

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157


INITIAL SECTION SELECTIONAssuming py = 275 N/mm2

S

Mpx

y

.

> =×307 5 10

275

6

= 1.118 × 106 mm3 = 1118 cm3

From steel tables (Appendix B ), suitable sections are:

1. 356 × 171 × 67 UB: Sx = 1210 cm3;2. 406 × 178 × 60 UB: Sx = 1190 cm3;3. 457 × 152 × 60 UB: Sx = 1280 cm3.

The above illustrates how steel beam sections are specified. For section 1, for instance, 356 ××××× 171 represents theserial size in the steel tables; 67 represents the mass per metre in kilograms; and UB stands for universal beam.

All the above sections give a value of plastic modulus about axis x–x, Sx, just greater than that required. Whicheverone is selected will depend on economic and engineering considerations. For instance, if lightness were the primaryconsideration, perhaps section 3 would be selected, which is also the strongest (largest Sx). However, if minimisingthe depth of the member were the main consideration then section 1 would be chosen. Let us choose the compromisecandidate, section 2.

CLASSIFICATION

Strength ClassificationBecause the flange thickness T = 12.8 mm (< 16 mm), then py = 275 N/mm2 (as assumed) from Table 4.3 and ε =(275/py)

1/2 = 1 (Table 4.4).

Section classificationb/T = 6.95 which is less than 9ε = 9. Hence from Table 4.4, flange is plastic. Also d /t = 46.2 which is less than 80ε= 80. Hence from Table 4.4, web is plastic. Therefore 406 × 178 × 60 UB section is class 1 plastic.

307.5

99

24

−24

−99(a) (b)

Fig. 4.12 (a) Bending moment and (b) shear force diagrams.

Because the structure is symmetrical RA = RB = 198/2 = 99 kN. The central bending moment, M, is

M

W

4= +

l l

ω 2

8

=

×+

×

48 104

15 108

2

= 120 + 187.5 = 307.5 kN m

Shear force and bending moment diagrams are shown in Fig. 4.12.


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158

Example 4.2 Selection of a beam section in grade S460 steel(BS 5950)

Repeat the above design in grade S460 steel.

INITIAL SECTION SELECTIONSince section is of grade S460 steel, assume py = 460 N/mm2

S

Mpx

y

> =×

. 307 5 10460

6

= 669 × 103 mm3 = 669 cm3

Suitable sections (with classifications) are:

1. 305 × 127 × 48 UB: Sx = 706 cm3, p y = 460, plastic;2. 305 × 165 × 46 UB: Sx = 723 cm3, p y = 460, compact;3. 356 × 171 × 45 UB: Sx = 774 cm3, Z = 687 cm3, p y = 460, semi-compact;4. 406 × 140 × 39 UB: Sx = 721 cm3, Z = 627 cm3, p y = 460, semi-compact.

Section 4 must be discounted. As it is semi-compact and Z is much less than 669 cm3 it fails in bending. Section 3is also semi-compact and Z is not sufficiently larger than 669 to take care of its own self weight. Section 2 looks thebest choice, being the light of the two remaining, and with the greater strength.

SHEAR STRENGTHAs d /t < 70ε, no shear buckling check is required. Shear capacity of section, Pv, is

Pv = 0.6pytD = 0.6 × 460 × 6.7 × 307.1

= 567.9 × 103 N = 567.9 kN

Now, as Fv(99 kN) < 0.6Pv = 340.7 kN (low shear load).

SHEAR STRENGTHAs d /t = 46.2 < 70ε, shear buckling need not be considered. Shear capacity of section, Pv, is

Pv = 0.6pyA v = 0.6pytD = 0.6 × 275 × 7.8 × 406.4

= 523 × 103 N = 523 kN

Now, as Fv(99 kN) < 0.6Pv = 0.6 × 523 = 314 kN (low shear load).

BENDING MOMENTMoment capacity of section, Mc, is

Mc = pyS = 275 × 1190 × 103

= 327 × 106 N mm = 327 kN m

≤ 1.2p y Z = 1.2 × 275 × 1060 × 103

= 349.8 × 106 N mm = 349.8 kN m OK

Moment M due to imposed loading = 307.5 kN m. Extra moment due to self weight, Msw, is

M sw = 1.4 × (60 × 9.81/103)

108

2

= 10.3 kN m

Total imposed moment M t = 307.5 + 10.3 = 317.8 kN m < 327 kN m. Hence proposed section is suitable.


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159


A

1.0 metre

450 kN total dead load270 kN total imposed load

Fig. 4.13

BENDING MOMENTMoment capacity of section subject to low shear load, Mc, is

Mc = pyS = 460 × 723 × 103

= 332.6 × 106 N mm = 332.6 kN m

≤ 1.2py Z = 1.2 × 460 × 648 × 103

= 357.7 × 106 N mm = 357.7 kN m OK

Moment M due to dead and imposed loading = 307.5 kN m. Extra moment due to self weight, Msw, is

M sw = 1.4 × (46 × 9.81/103)

108

2

= 7.9 kN m

Total imposed moment M t = 307.5 + 7.9 = 315.4 kN m, which is less than the section’s moment of resistance Mc =332.6 kN m. This section is satisfactory.


Example 4.3 Selection of a cantilever beam section (BS 5950)A proposed cantilever beam 1 m long is to be built into a concrete wall as shown in Fig. 4.13. It supportscharacteristic dead and imposed loading of 450 kN/m and 270 kN/m respectively. Select a suitable UB section inS275 steel to satisfy bending and shear criteria only.

4.8.7 HIGH SHEAR AND MOMENT CAPACITYWhen the shear load is high, i.e. Fv > 0.6Pv, themoment-carrying capacity of the section is reduced.This is because the web cannot take the full tensileor compressive stress associated with the bendingmoment as well as a sustained substantial shearstress due to the shear force. Thus, according toclause 4.2.5.3 of BS 5950, the moment capacity ofUB and UC sections, Mc, should be calculated asfollows:

For class 1 plastic and compact sections

Mc = py(S − ρSv) (4.11)

For class 3 semi-compact sections

Mc = py(S − ρSv/1.5) or alternatively

Mc = py(Seff − ρSv) (4.12)


Mc = py(Zeff − ρSv /1.5) (4.13)

where ρ = [2(Fv/Pv) − 1]2 and Sv for sections withequal flanges, is the plastic modulus of the sheararea of section equal to tD2/4. The other symbolsare as previously defined in section 4.8.6.

Note the effect of the ρ factor is to reduce themoment-carrying capacity of the web as the shearload rises from 50 to 100% of the web’s shearcapacity. However, the resulting reduction in mo-ment capacity is negligible when Fv < 0.6Pv.

4.8.8 DEFLECTIONA check should be carried out on the maximumdeflection of the beam due to the most adverserealistic combination of unfactored imposed

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160

Example 4.3 continuedDESIGN BENDING MOMENT AND SHEAR FORCEShear force Fv at A is

(450 × 1.4) + (270 × 1.6) = 1062 kN

Bending Moment M at A is

Wb2

1062 12

531

=×

= kN m


S

Mpx

y

mm cm> =×

= × =

531 10

2751931 10 1931

63 3 3

Suitable sections (with classifications) are:

1. 457 × 191 × 89 UB: Sx = 2010 cm3, py = 265, plastic;2. 533 × 210 × 82 UB: Sx = 2060 cm3, py = 275, plastic.

SHEAR STRENGTHShear capacity of 533 × 210 × 82 UB section, Pv, is

P v = 0.6pytD = 0.6 × 275 × 9.6 × 528.3

= 837 × 103 N = 837 kN < 1062 kN Not OK

In this case, because the length of the cantilever is so short, the selection of section will be determined from theshear strength, which is more critical than bending. Try a new section:

610 × 229 × 113 UB: Sx = 3290 cm3, py = 265, plastic

P v = 0.6 × 265 × 607.3 × 11.2 = 1081 × 103 N

= 1081 kN OK but high shear load

BENDING MOMENTFrom above

ρ = [2(Fv/Pv) − 1]2 = [2(1062/1081) − 1]2 = 0.93

Mc = p y(Sx − ρS v)

= 265(3290 × 103 − 0.93[11.2 × 607.32/4] )

= 617 × 106 = 617 kN m

≤ 1.2pyZ = 1.2 × 265 × 2880 × 103

= 915.8 × 106 = 915.8 kN m OK

Msw = 1.4 × (113 × 9.81/103)

12

2 = 0.8 kN m

Mt = M + Msw = 531 + 0.8 = 532 kN m < Mc

This section is satisfactory.

serviceability loading. In BS 5950 this is coveredby clause 2.5.2 and Table 8, part of which is reprod-uced as Table 4.5. Table 8 outlines recommendedlimits to these deflections, compliance with which

should avoid significant damage to the structureand finishes.

Calculation of deflections from first principleshas to be done using the area-moment method,

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161


Table 4.5 Suggested vertical deflection limitson beams due to imposed load (based onTable 8, BS 5950)

Cantilevers Length /180Beams carrying plaster or other Span/360

brittle finishOther beams (except purlins and Span/200

sheeting rails)

Macaulay’s method, or some other similar approach,a subject which is beyond the scope of this book.

The reader is referred to a suitable structuralanalysis text for more detail on this subject. How-ever, many calculations of deflection are carriedout using formulae for standard cases, which canbe combined as necessary to give the answer formore complicated situations. Figure 4.14 summar-ises some of the more useful formulae.

A

AC w/unit length

B A BC

W point load

span � span �

length � length �

B A B

δC = 5wl 4

384EIδC = Wl 3

48EI

δB = wl 4

8EI

W

δB = Wl 3

3EI

Fig. 4.14 Deflections for standard cases. E = elastic modulus of steel (205 kN/mm2) and I = second moment of area (x–x)of section.

Example 4.4 Deflection checks on steel beams (BS 5950)Carry out a deflection check for Examples 4.1–4.3 above.

FOR EXAMPLE 4.1

δ

ωC = +

5384 48

4 3b bEI

WEI

=

× ×× × × ×

+×

× × × ×− −

5 5 10

384 205 10 21500 1030 10

48 205 10 21500 10

4

6 8

3

6 8

= 0.0148 + 0.0142 = 0.029 m = 29 mm

From Table 4.5, the recommended maximum deflection for beams carrying plaster is span/360 which equals 10000/360= 27.8 mm, and for other beams span/200 = 10000/200 = 50 mm. Therefore if the beam was carrying a plaster finishone might consider choosing a larger section.

FOR EXAMPLE 4.2

δ

ωC = +

5384 48

4 3b bEI

WEI

=

× ×× × × ×

+×

× × × ×− −

5 5 10

384 205 10 9950 1030 10

48 205 10 9950 10

4

6 8

3

6 8

= 0.0319 + 0.0306 = 0.0625 m = 62.5 mm

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162

4.8.9 WEB BEARING AND WEB BUCKLINGClause 4.5 of BS 5950 covers all aspects ofweb bearing, web buckling and stiffener design.Usually most critical at the position of a supportor concentrated load is the problem of webbuckling. The buckling resistance of a web isobtained via the web bearing capacity as discussedbelow.

4.8.9.1 Web bearingFigure 4.15 (a) illustrates how concentrated loadsare transmitted through the flange/web connectionin the span, and at supports when the distance tothe end of the member from the end of the stiffbearing is zero. According to Clause 4.5.2 of BS5950, the bearing resistance Pbw is given by:

Pbw = (b1 + nk)tpyw (4.15)

whereb1 is the stiff bearing lengthn is as shown in Figure 4.15(a): n = 5 except

at the end of a member and n = 2 + 0.6be/k≤ 5 at the end of the member

be is the distance to the end of the member fromthe end of the stiff bearing (Fig. 4.15(b))

k = (T + r) for rolled I- or H-sectionsT is the thickness of the flanget is the web thicknesspyw is the design strength of the web

This is essentially a simple check which ensuresthat the stress at the critical point on the flange/web connection does not exceed the strength ofthe steel.

It should also be checked that the contact stressesbetween load or support and flange do not exceedpy.

4.8.9.2 Web bucklingAccording to clause 4.5.3.1 of BS 5950, provided thedistance αe from the concentrated load or reactionto the nearer end of the member is at least 0.7d,and if the flange through which the load or reac-tion is applied is effectively restrained against both

(a) rotation relative to the web(b) lateral movement relative to the other flange

(Fig. 4.16 ),

the buckling resistance of an unstiffened web isgiven by

P

t

b nk dPx bw

( )=

+25

1

ε(4.16)

The reader is referred to Appendix C for thebackground and derivation of this equation.

Alternatively, when αe < 0.7d, the buckling re-sistance of an unstiffened web is given by

P

dd

t

b nk dPx

e

1

bw1.4 (

.

)=

++

α ε0 7 25(4.17)

If the flange is not restrained against rotationand/or lateral movement the buckling resistance ofthe web is reduced to Pxr, given by

P

dL

PxrE

x .

=0 7

(4.18)

in which LE is the effective length of the web de-termined in accordance with Table 22 of BS 5950.

Example 4.4 continuedThe same recommended limits from Table 4.5 apply as above. This grade S460 beam therefore fails the deflection test.This partly explains why the higher strength steel beams are not particularly popular.

FOR EXAMPLE 4.3

δ

ωB

= =×

× × × × −

b4 4

6 88270 1

8 205 10 87400 10EI

= 0.00019 m = 0.19 mm

The recommended limit from Table 4.5 is

Length/180 = 5.6 mm.

So deflection is no problem for this particular beam.

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163


(a)Roo

t rad

ius

r

Fla

nge

thic

knes

s T

1:2.5

1:2b1 2k

2.5k b1 2.5k

Effective contactarea 2(r + T )

b1

be

αe

(b)

s

s

b1 + nk s

sb1 + nk

Fig. 4.15 Web bearing.

(a) (b)

Fig. 4.16

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164

Example 4.5 Checks on web bearing and buckling for steel beams(BS 5950)

Check web bearing and buckling for Example 4.1, assuming the beam sits on 100 mm bearings at each end.

WEB BEARING AT SUPPORTSPbw = (b1 + nk)tpyw

= (100 + 2 × 23)7.8 × 275

= 313 × 103 N = 313 kN > 99 kN OK

wherek = T + r = 12.8 + 10.2 = 23 mmn = 2 + 0.6be/k = 2 (since be = 0)

CONTACT STRESS AT SUPPORTSPcs = (b1 × 2(r +T ))py = (100 × 46) × 275

= 1265 × 103 N = 1265 kN > 99 kN OK

WEB BUCKLING AT SUPPORTSince αe (= 50 mm) < 0.7d = 0.7 × 360.5 = 252 mm, buckling resistance of the web is

P

dd

t

b nk dPx

e

1bw1.4 (

.

) =

++

α ε0 7 25

=

+×

×× ×

+ ×=

. .

.

( ) .

50 2521 4 360 5

25 1 7 8

100 2 23 360 5313 159 kN > 99 kN OK

So no web stiffeners are required at supports.A web check at the concentrated load should also be carried out, but readers can confirm that this aspect is not

critical in this case.

200 kN/m dead load100 kN/m imposed load

A BC

5 metres

Fig. 4.17

Example 4.6 Design of a steel beam with web stiffeners (BS 5950)A simply supported beam is to span 5 metres and support uniformly distributed characteristic dead and imposed loadsof 200 kN/m and 100 kN/m respectively. The beam sits on 150 mm long bearings at supports, and both flanges arelaterally and torsionally restrained (Fig. 4.17 ). Select a suitable UB section to satisfy bending, shear and deflectioncriteria. Also check web bearing and buckling at supports, and design stiffeners if they are required.

carrying stiffeners can be locally welded betweenthe flanges and the web. Clause 4.5 of BS 5950gives guidance on the design of such stiffeners andthe following example illustrates this.

4.8.10 STIFFENER DESIGNIf it is found that the web fails in buckling or bear-ing, it is not always necessary to select anothersection; larger supports can be designed, or load-

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165


Example 4.6 continuedDESIGN SHEAR FORCE AND BENDING MOMENTFactored loading = (200 × 1.4) + (100 × 1.6) = 440 kN/mReactionsRA = RB = 440 × 5 × 0.5 = 1100 kN

Bending moment, M

= =

×=

ωb2 2

8440 5

81375 kN m


S

Mpx

y

3 3mm cm

> =×

= × =1375 10

2755000 10 5000

63

From Appendix B, suitable sections (with classifications) are:

1. 610 × 305 × 179 UB: Sx = 5520 cm3, p y = 265 plastic;2. 686 × 254 × 170 UB: Sx = 5620 cm3, p y = 265 plastic;3. 762 × 267 × 173 UB: Sx = 6200 cm3, p y = 265 plastic;4. 838 × 292 × 176 UB: Sx = 6810 cm3, p y = 265 plastic.

Section 2 looks like a good compromise; it is the lightest section, and its depth is not excessive.

SHEAR STRENGTH

Pv = 0.6pytD = 0.6 × 265 × 14.5 × 692.9

= 1597 × 103 N = 1597 kN > 1100 kN OK

However as Fv (= 1100 kN) > 0.6Pv (= 958 kN), high shear load.

BENDING MOMENTAs the shear force is zero at the centre, the point of maximum bending moment, Mc is obtained from

Mc = pyS = 265 × 5620 × 103

= 1489.3 × 106 N mm = 1489.3 kN m

≤ 1.2pyZ = 1.2 × 265 × 4910 × 103

= 1561.3 × 106 N mm = 1561.3 kN m OK

Msw = 1.4 × ( . / )170 9 81103×

58

2

= 7.3 kN m

Total moment Mt = 1375 + 7.3 = 1382.3 kN m < 1489.3 kN m. OK

DEFLECTIONCalculated deflection, δC, is

δ

ωC = =

× ×× × × × −

5384

5 100 5384 205 10 170000 10

4 4

6 8

bEI

= 0.0023 m = 2.3 mm

Maximum recommended deflection limit for a beam carrying plaster from Table 4.5 = span /360 = 13.8 mm OK

9780415467193_C04 9/3/09, 1:24 PM165


166

WEB BEARINGk = T + r = 23.7 + 15.2 = 38.9 mm

n = 2 + 0.6be /k = 2 (since be = 0)

Pbw = (b1 + nk)tpyw

= (150 + 2 × 38.9)14.5 × 275

= 908 × 103 N = 908 kN

The web’s bearing resistance Pbw (= 908 kN) < RA (= 1100 kN), and so load-carrying stiffeners are required. BS 5950stipulates that load-carrying stiffeners should be checked for both bearing and buckling.

STIFFENER BEARING CHECKLet us propose 12 mm thick stiffeners each side of the web and welded continuously to it. As the width of sectionB = 255.8 mm, the stiffener outstand is effectively limited to 120 mm.

The actual area of stiffener in contact with the flange, if a 15 mm fillet is cut out for the root radius, A s,net =(120 − 15)2 × 12 = 2520 mm2

The bearing capacity of the stiffener, Ps, is given by (clause 4.5.2.2)

Ps = A s,net py = 2520 × 275

= 693 × 103 N > external reaction = (RA − Pbw = 1100 − 908 =) 192 kN

Hence the stiffener provided is adequate for web bearing.

WEB BUCKLINGSince αe (= 75 mm) < 0.7d = 0.7 × 692.9 = 485 mm, buckling resistance of the web is

P

dd

t

b nk dPx

e

1bw

(

.. )

=+

+α ε0 7

1 425

=

+×

×× ×

+ ×=

. .

.

( . ) . .

75 4851 4 692 9

25 1 14 5

150 2 38 9 692 9908 478 3 1100 kN kN�

Therefore, web stiffeners capable of resisting an external (buckling) load, Fx, of Fx = (Fv − Px) = 1100 − 478.3 = 621.7 kNare required.

STIFFENER BUCKLING CHECKThe buckling resistance of a stiffener, Pxs, is given by

Pxs = A s pc

The plan of the web and stiffener at the support position is shown in Fig. 4.18. Note that a length of web on each sideof the centre line of the stiffener not exceeding 15 times the web thickness should be included in calculating thebuckling resistance (clause 4.5.3.3). Hence, second moment of area (Is) of the effective section (shown cross-hatchedin Fig. 4.18) (based on bd 3/12) for stiffener buckling about the z–z axis is

Is mm=

× + ++

+ ×= ×

( . )

( . ) . .

12 120 120 14 512

217 5 69 14 512

16 56 103 3

6 4

Effective area of buckling section, A s, is

A s = 12 × (120 + 120 + 14.5) + (217.5 + 69) × 14.5

= 7208 mm2


9780415467193_C04 9/3/09, 1:24 PM166

167


they provide at the beam supports, the higher theload required to make the beam twist sideways.This effect is taken into account by using the con-cept of ‘effective length’, as discussed below.

4.8.11.1 Effective lengthThe concept of effective length is introduced inclause 4.3.5 of BS 5950. For a beam supported atits ends only, with no intermediate lateral restraint,and standard restraint conditions at supports, theeffective length is equal to the actual length be-tween supports. When a greater degree of lateraland torsional restraint is provided at supports, theeffective length is less than the actual length, andvice versa. The effective length appropriate todifferent end restraint conditions is specified inTable 13 of BS 5950, reproduced as Table 4.6, and

4.8.11 LATERAL TORSIONAL BUCKLINGIf the loaded flange of a beam is not effectivelyrestrained against lateral movement relative to theother flange, by a concrete floor fixed to the beam,for instance, and against rotation relative to theweb, by web cleats or fin plates, for example, it ispossible for the beam to twist sideways under aload less than that which would cause the beam tofail in bending, shear or deflection. This is calledlateral torsional buckling which is covered byClause 4.3 of BS 5950. It is illustrated by Fig. 4.19for a cantilever, but readers can experiment with thisphenomenon very easily using a scale rule or rulerloaded on its edge. They will see that although theproblem occurs more readily with a cantilever, italso applies to beams supported at each end. Keenexperimenters will also find that the more restraint


Fig. 4.18 Plan of web and stiffener.

Outline of flangex

69 120

z

t = 14.5

Stiffenerx

120

12

Web

15t = 217.5

z

Radius of gyration r = (Is/A s)0.5

= (16.56 × 106/7208)0.5

= 47.9 mm

According to Clause 4.5.3.3 of BS 5950, the effective length (L E) of load carrying stiffeners when the compressionflange is laterally restrained = 0.7L, where L = length of stiffener (= d )

= 0.7 × 615.1

= 430.6 mm

λ = LE /r = 430.6/47.9 = 9

Then from Table 24(c) (Table 4.14 ) pc = p y = 275 N/mm2

Then Pxs = A spc = 7208 × 275

= 1982 × 103 N

= 1982 kN > Fx = 621.7 kN

Hence the stiffener is also adequate for web buckling.

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168

Root of cantileverDestabilising load

End of cantilever

Normal load

Normal load

Support

Destabilising load

Support

(a) (b)

Fig. 4.19 Lateral torsional buckling: (a) cantilever beam; ( b) simply supported beam.

Table 4.6 Effective length, LE, for beams (Table 13, BS 5950)

Conditions of restraint at supports Loading conditions

Normal Destab.

Comp. flange laterally restrainedNominal torsional restraint against rotation about longitudinal axisBoth flanges fully restrained against rotation on plan 0.7L LT 0.85L LT

Compression flange fully restrained against rotation on plan 0.75L LT 0.9L LT

Both flanges partially restrained against rotation on plan 0.8L LT 0.95L LT

Compression flange partially restrained against rotation on plan 0.85L LT 1.0LBoth flanges free to rotate on plan 1.0L LT 1.2L LT

Comp. flange laterally unrestrainedBoth flanges free to rotate on planPartial torsional restraint against rotation about longitudinal 1.0L LT + 2D 1.2LLT + 2D

axis provided by connection of bottom flange to supportsPartial torsional restraint against rotation about longitudinal 1.2L LT + 2D 1.4LLT + 2D

axis provided only by pressure of bottom flange onto supports

illustrated in Fig. 4.20. The table gives differentvalues of effective length depending on whetherthe load is normal or destabilising. This is perhapsbest explained using Fig. 4.19, in which it can read-ily be seen that a load applied to the top of thebeam will cause it to twist further, thus worseningthe situation. If the load is applied below thecentroid of the section, however, it has a slightlyrestorative effect, and is conservatively assumed tobe normal.

Determining the effective length for real beams,when it is difficult to define the actual conditionsof restraint, and whether the load is normal ordestabilising, is perhaps one of the greatest prob-lems in the design of steelwork. It is an aspect that

was amended in the 1990 edition of BS 5950 andonce again in the 2000 edition in order to relatemore to practical circumstances.

4.8.11.2 Lateral torsional buckling resistanceChecking of lateral torsional buckling for rolledUB sections can be carried out in two differentways. Firstly, there is a slightly conservative, butquite simple check in clause 4.3.7 of BS 5950,which only applies to equal flanged rolled sections.The approach is similar to that for struts (section4.9.1), and involves determining the bendingstrength for the section, pb, from Table 20 of BS5950, reproduced as Table 4.7, via the slendernessvalue (βw)0.5LE/ry and D/T.

9780415467193_C04 9/3/09, 1:25 PM168

169

Fig. 4.20 Restraint conditions.

Columnassumed to betorsionally stiff

Bolts preventuplift at flangeonly

DETAIL AT SUPPORTS

Compression flange laterally restrainedNominal torsional restraint againstrotation about longitudinal axisBoth flanges fully restrained againstrotation on plan

Compression flange laterally restrainedNominal torsional restraint againstrotation about longitudinal axisBoth flanges free to rotate on plan

Compression flange laterally unrestrainedBoth flanges free to rotate on planPartial torsional restraint against rotationabout longitudinal axis provided byconnection of bottom flange to support

PLAN ON BEAMKEY

Tension flange restraintCompression flange restraint


Table 4.7 Bending strength p b (N/mm2) for rolled sections with equal flanges (Table 20, BS 5950)

1) Grade S275 steel ≤ 16 mm ( py = 275 N/mm2)

(βw)0.5LE/ry D/T

5 10 15 20 25 30 35 40 45 50

30 275 275 275 275 275 275 275 275 275 27535 275 275 275 275 275 275 275 275 275 27540 275 275 275 275 274 273 272 272 272 27245 275 275 269 266 264 263 263 262 262 26250 275 269 261 257 255 253 253 252 252 25155 275 263 254 248 246 244 243 242 241 24160 275 258 246 240 236 234 233 232 231 23065 275 252 239 232 227 224 223 221 221 22070 274 247 232 223 218 215 213 211 210 20975 271 242 225 215 209 206 203 201 200 19980 268 237 219 208 201 196 193 191 190 18985 265 233 213 200 193 188 184 182 180 17990 262 228 207 193 185 179 175 173 171 16995 260 224 201 186 177 171 167 164 162 160

100 257 219 195 180 170 164 159 156 153 152105 254 215 190 174 163 156 151 148 146 144110 252 211 185 168 157 150 144 141 138 136115 250 207 180 162 151 143 138 134 131 129120 247 204 175 157 145 137 132 128 125 123125 245 200 171 152 140 132 126 122 119 116130 242 196 167 147 135 126 120 116 113 111135 240 193 162 143 130 121 115 111 108 106

9780415467193_C04 9/3/09, 1:25 PM169


170


(βw)0.5LE/ry D/T

5 10 15 20 25 30 35 40 45 50

140 238 190 159 139 126 117 111 106 103 101145 236 186 155 135 122 113 106 102 99 96150 233 183 151 131 118 109 102 98 95 92155 231 180 148 127 114 105 99 94 91 88160 229 177 144 124 111 101 95 90 87 84165 227 174 141 121 107 98 92 87 84 81170 225 171 138 118 104 95 89 84 81 78175 223 169 135 115 101 92 86 81 78 75180 221 166 133 112 99 89 83 78 75 72185 219 163 130 109 96 87 80 76 72 70190 217 161 127 107 93 84 78 73 70 67195 215 158 125 104 91 82 76 71 68 65200 213 156 122 102 89 80 74 69 65 63210 209 151 118 98 85 76 70 65 62 59220 206 147 114 94 81 72 66 62 58 55230 202 143 110 90 78 69 63 58 55 52240 199 139 106 87 74 66 60 56 52 50250 195 135 103 84 72 63 57 53 50 47

2) Grade S275 steel > 16 mm ≤ 40 mm ( py = 265 N/mm2)

30 265 265 265 265 265 265 265 265 265 26535 265 265 265 265 265 265 265 265 265 26540 265 265 265 265 265 264 264 264 263 26345 265 265 261 258 256 255 254 254 254 25450 265 261 253 249 247 246 245 244 244 24455 265 255 246 241 238 236 235 235 234 23460 265 250 239 233 229 227 226 225 224 22465 265 245 232 225 221 218 216 215 214 21470 265 240 225 217 212 209 207 205 204 20475 263 235 219 210 204 200 198 196 195 19480 260 230 213 202 196 191 189 187 185 18485 257 226 207 195 188 183 180 178 176 17590 254 222 201 188 180 175 171 169 167 16695 252 217 196 182 173 167 163 160 158 157

100 249 213 190 176 166 160 156 153 150 149105 247 209 185 170 160 153 148 145 143 141110 244 206 180 164 154 147 142 138 136 134115 242 202 176 159 148 140 135 132 129 127120 240 198 171 154 142 135 129 125 123 121125 237 195 167 149 137 129 124 120 117 115130 235 191 163 144 132 124 119 114 111 109135 233 188 159 140 128 119 114 109 106 104140 231 185 155 136 124 115 109 105 102 99145 229 182 152 132 120 111 105 101 97 95150 227 179 148 129 116 107 101 97 93 91155 225 176 145 125 112 103 97 93 89 87160 223 173 142 122 109 100 94 89 86 83165 221 170 139 119 106 97 91 86 83 80170 219 167 136 116 103 94 88 83 80 77175 217 165 133 113 100 91 85 80 77 74180 215 162 130 110 97 88 82 77 74 71185 213 160 128 108 95 86 79 75 71 69

9780415467193_C04 9/3/09, 1:25 PM170

171



(βw)0.5LE/ry D/T

5 10 15 20 25 30 35 40 45 50

190 211 157 125 105 92 83 77 73 69 66195 209 155 123 103 90 81 75 70 67 64200 207 153 120 101 88 79 73 68 65 62210 204 148 116 96 84 75 69 64 61 58220 200 144 112 93 80 71 65 61 58 55230 197 140 108 89 77 68 62 58 54 52240 194 136 104 86 74 65 59 55 52 49250 190 132 101 83 71 63 57 52 49 47

For class 1 plastic or class 2 compact sections,the buckling resistance moment, Mb, is obtainedfrom

Mb = pbSx (4.19)

For class 3 semi-compact sections, Mb, is givenby

Mb = pbZx (4.20)

whereSx plastic modulus about the major axisZx elastic modulus about the major axisβw a ratio equal to 1 for class 1 plastic or class

2 compact sections and Zx/Sx for class 3semi-compact sections

D depth of the sectionry radius of gyration about the y–y axisT flange thickness

Example 4.7 Design of a laterally unrestrained steel beam(simple method) (BS 5950)

Assuming the beam in Example 4.1 is not laterally restrained, determine whether the selected section is suitable, andif not, select one which is. Assume the compression flange is laterally restrained and the beam fully restrained againsttorsion at supports, but both flanges are free to rotate on plan. The loading is normal.

EFFECTIVE LENGTHSince beam is pinned at both ends, from Table 4.6, LE = 1.0L LT = 10 m

BUCKLING RESISTANCEUsing the conservative approach of clause 4.3.7

( ) ( . )

. . .βw

E

y

0 5 0 51 01000039 7

252Lr = =

DT

.

. .= =

406 412 8

31 75

From Table 4.7, pb = 60 N/mm2 (approx).

Mb = pbSx = 60 × 1190 × 103

= 71.4 × 106 N mm = 71.4 kN m << 317.8 kN m

As the buckling resistance moment is much less than the actual imposed moment, this beam would fail by lateraltorsional buckling. Using trial and error, a more suitable section can be found.

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172

Table 4.8 Slenderness factor ν for beams withequal flanges (based on Table 19, BS 5950)

λx

N = 0.5

0.5 1.001.0 0.991.5 0.972.0 0.962.5 0.933.0 0.913.5 0.894.0 0.864.5 0.845.0 0.825.5 0.796.0 0.776.5 0.757.0 0.737.5 0.728.0 0.708.5 0.689.0 0.679.5 0.6510.0 0.6411.0 0.6112.0 0.5913.0 0.5714.0 0.5515.0 0.5316.0 0.5217.0 0.5018.0 0.4919.0 0.4820.0 0.47

4.8.11.3 Equivalent slenderness anduniform moment factors

The more rigorous approach for calculating valuesof Mb is covered by clauses 4.3.6.2 to 4.3.6.9 ofBS 5950. It involves calculating an equivalent slen-derness ratio, λLT, given by

λLT = uνλ βw (4.21)

in which

λ =

Lr

E

y

whereLE effective length for lateral torsional bucklingry radius of gyration about the minor axisu buckling parameter = 0.9 for rolled I- and

H-sectionsν slenderness factor from Table 19 of BS

5950, part of which is reproduced asTable 4.8, given in terms of λ /x in whichx torsional index = D/T where D is thedepth of the section and T is the flangethickness

βw = 1.0 for class 1 plastic or class 2 compactsections;= Zx/Sx for class 3 semi-compact sections ifMb = p bZ x;= Sx,eff/Sx for class 3 semi-compact sections ifMb = pbSx,eff ;= Zx,eff/Sx for class 4 slender cross-sections.

Knowing λLT and the design strength, py, thebending strength, pb, is then read from Table 16for rolled sections, reproduced as Table 4.9. The

Try 305 × 305 × 137 UC; p y = 265, plastic.

( ) ( . )

. . .βw

E

y

0 5 0 51 01000078 2

128Lr = =

DT

..

.= =320 521 7

14 8

Then from Table 4.7, pb = 165.7 N/mm2

Mb = pbSx = 165.7 × 2300 × 103 = 381 × 106 N mm = 381 kN m

M sw = 1.4 ( . / )137 9 8110

108

32

× = 23.5 kN m

Total imposed moment M t = 307.5 + 23.5 = 331 kN m < 381 kN m OKSo this beam, actually a column section, is suitable. Readers may like to check that there are not any lighter UB

sections. Because the column has a greater ry, it is laterally stiffer than a UB section of the same weight and is moresuitable than a beam section in this particular situation.


9780415467193_C04 9/3/09, 1:26 PM172

173


Table 4.9 Bending strength pb (N/mm2) for rolled sections (Table 16, BS 5950)

λLT Steel grade and design strength, py (N/mm2)

S275 S355 S460

235 245 255 265 275 315 325 335 345 355 400 410 430 440 460

25 235 245 255 265 275 315 325 335 345 355 400 410 430 440 46030 235 245 255 265 275 315 325 335 345 355 395 403 421 429 44635 235 245 255 265 273 307 316 324 332 341 378 386 402 410 42640 229 238 246 254 262 294 302 309 317 325 359 367 382 389 40445 219 227 235 242 250 280 287 294 302 309 340 347 361 367 38150 210 217 224 231 238 265 272 279 285 292 320 326 338 344 35655 199 206 213 219 226 251 257 263 268 274 299 305 315 320 33060 189 195 201 207 213 236 241 246 251 257 278 283 292 296 30465 179 185 190 196 201 221 225 230 234 239 257 261 269 272 27970 169 174 179 184 188 206 210 214 218 222 237 241 247 250 25675 159 164 168 172 176 192 195 199 202 205 219 221 226 229 23480 150 154 158 161 165 178 181 184 187 190 201 203 208 210 21485 140 144 147 151 154 165 168 170 173 175 185 187 190 192 19590 132 135 138 141 144 153 156 158 160 162 170 172 175 176 17995 124 126 129 131 134 143 144 146 148 150 157 158 161 162 164

100 116 118 121 123 125 132 134 136 137 139 145 146 148 149 151105 109 111 113 115 117 123 125 126 128 129 134 135 137 138 140110 102 104 106 107 109 115 116 117 119 120 124 125 127 128 129115 96 97 99 101 102 107 108 109 110 111 115 116 118 118 120120 90 91 93 94 96 100 101 102 103 104 107 108 109 110 111125 85 86 87 89 90 94 95 96 96 97 100 101 102 103 104130 80 81 82 83 84 88 89 90 90 91 94 94 95 96 97135 75 76 77 78 79 83 83 84 85 85 88 88 89 90 90140 71 72 73 74 75 78 78 79 80 80 82 83 84 84 85145 67 68 69 70 71 73 74 74 75 75 77 78 79 79 80150 64 64 65 66 67 69 70 70 71 71 73 73 74 74 75155 60 61 62 62 63 65 66 66 67 67 69 69 70 70 71160 57 58 59 59 60 62 62 63 63 63 65 65 66 66 67165 54 55 56 56 57 59 59 59 60 60 61 62 62 62 63170 52 52 53 53 54 56 56 56 57 57 58 58 59 59 60175 49 50 50 51 51 53 53 53 54 54 55 55 56 56 56180 47 47 48 48 49 50 51 51 51 51 52 53 53 53 54185 45 45 46 46 46 48 48 48 49 49 50 50 50 51 51190 43 43 44 44 44 46 46 46 46 47 48 48 48 48 48195 41 41 42 42 42 43 44 44 44 44 45 45 46 46 46200 39 39 40 40 40 42 42 42 42 42 43 43 44 44 44210 36 36 37 37 37 38 38 38 39 39 39 40 40 40 40220 33 33 34 34 34 35 35 35 35 36 36 36 37 37 37230 31 31 31 31 31 32 32 33 33 33 33 33 34 34 34240 28 29 29 29 29 30 30 30 30 30 31 31 31 31 31250 26 27 27 27 27 28 28 28 28 28 29 29 29 29 29λL0 37.1 36.3 35.6 35 34.3 32.1 31.6 31.1 30.6 30.2 28.4 28.1 27.4 27.1 26.5

9780415467193_C04 9/3/09, 1:27 PM173


174

Fig. 4.21 Standard case for buckling.

M M

M M

L

Fig. 4.22 Load cases less susceptible to buckling.

W

L

M = WL/8

(a)

M βM

βM

M

L

(b)

buckling resistance moment, Mb, is obtained fromthe following:

For class 1 plastic or cass 2 compact sections

Mb = pbSx (4.22)

For class 3 semi-compact sections, Mb, is givenby

Mb = pbZx (4.23)

or alternatively Mb = pbSx,eff (4.24)


Mb = pbZx,eff (4.25)

whereSx plastic modulus about the major axisZx elastic modulus about the major axisSx,eff effective plastic modulus about the major

axis (see clause 3.5.6)Zx,eff effective section modulus about the major

axis (see clause 3.6.2)

This value of the buckling moment assumes thebeam is acted on by a uniform, single curvaturemoment (Fig. 4.21), which is the most severearrangement in terms of lateral stability. However,where the beam is acted on by variable moments

(Fig. 4.22(a)) or unequal end moments (Fig.4.22(b)), the maximum moment about the majoraxis, Mx, must satisfy the following conditions:

Mx ≤ Mb/mLT and Mx ≤ Mcx (4.26)

wheremLT equivalent uniform moment factor for

lateral torsional buckling read from Table18 of BS 5950, reproduced as Table 4.10.For the destabilising loading conditionmLT = 1

Mcx moment capacity of the section about themajor axis

Example 4.8 Design of a laterally unrestrained beam – rigorous method(BS 5950)

Repeat Example 4.7 using the more rigorous method.As previously noted the 305 × 305 × 137 UC section is class 1 plastic has a design strength, py = 265 N/mm2. From

steel tables (Appendix B), r y = 78.2 mm, D = 320.5 mm and T = 21.7 mm. The slenderness ratio, λ, is

λ

. .= = =

Lr

E

y

1000078 2

127 9

λx

.

. / . .= =

127 9320 5 21 7

8 7

From Table 4.8, ν = 0.678. Since the section is class 1 plastic, β w = 1.0 and the equivalent slenderness ratio, λ LT, is

λ LT = uνλ βw = 0.9 × 0.678 × 127.9 × 1 0. = 78

9780415467193_C04 11/3/09, 11:07 AM174

175

Segments with end moments only β

1.00.90.80.70.60.50.40.30.20.10.0

−0.1−0.2−0.3−0.4−0.5−0.6−0.7−0.8−0.9−1.0

mLT

1.000.960.920.880.840.800.760.720.680.640.600.560.520.480.460.440.440.440.440.440.44

β negative

x

M

x Lateral restraint

x

βM

LLT

x

M

x

βM

LLT

x

x

x

x

MLLT

LLT

βM

βM

x

x

x

x

M

Specific cases (no intermediate lateral restraints)

LLT LLT

x x x x

mLT = 0.850 mLT = 0.925

x x x x

LLT LLT

mLT = 0.925 mLT = 0.744

β positive

Table 4.10 Equivalent uniform moment factor mLT for lateraltorsional buckling (based on Table 18, BS 5950)


From Table 4.9, p b = 165.4 N/mm2, hence

Mb = pbSx = 165.4 × 2300 × 103

= 380.4 × 106 N mm = 380.4 kN m

m LT from Table 4.10 = 0.89 (approx.), by interpolation between 0.85 and 0.925

Mm

b

LT

.

.=

380 40 89

= 427.4 kN m < Mcx (= p y Sx = 265 × 2300 × 10−3) = 609.5 kN m

This gives, in this case, a buckling moment approximately 10% greater than in Example 4.7. This may enable a lightermember to be selected, but for rolled sections it may not be really worth the additional effort.


9780415467193_C04 9/3/09, 1:27 PM175


176

Table 4.11 Effective length, LE, for cantilevers without intermediate constraint (based on Table 14,BS 5950)

Restraint conditions at supports Restraint conditons at tip Loading conditions

Normal Destabilising

Restrained laterally, torsionally 1) Free 0.8L 1.4Land against rotation on plan 2) Lateral restraint to top flange 0.7L 1.4L

3) Torsional restraint 0.6L 0.6L4) Lateral and torsional restraint 0.5L 0.5L

L

4.8.11.4 Cantilever beamsFor cantilevers, the effective length is given in cleardiagrammatical form in Table 14 of BS 5950, partof which is reproduced as Table 4.11.

4.8.11.5 Summary of design proceduresThe two alternative methods for checking lateraltorsional buckling of beams can be summarised asfollows.

Conservative method

1. Calculate the design shear force, Fv, and bendingmoment, Mx, at critical points along the beam.

2. Select and classify UB or UC section.3. Check shear and bending capacity of section. If

unsatisfactory return to (2).4. Determine the effective length of the beam, LE,

using Table 4.6.

Example 4.9 Checking for lateral instability in a cantilever steel beam(BS 5950)

Continue Example 4.3 to determine whether the cantilever is laterally stable. Assume the load is destabilising.From Table 4.11, LE = 1.4L = 1.4 mFor 610 × 229 × 113 UB

λ

. .= = =

Lr

E

y

140048 8

28 7

λx

.

. / . .= =

28 7607 3 17 3

0 82 ⇒ ν = 0.99 (Table 4.8)

λ LT = uνλ βw = 0.9 × 0.99 × 28.7 × 1 0. = 28.4

From Table 4.9, pb = 265 N/mm2 = py, and so lateral torsional buckling is not a problem with this cantilever.

9780415467193_C04 9/3/09, 1:28 PM176

177

5. Determine Sx, ry, D and T from steel tables(Appendix B).

6. Calculate the slenderness value, λ = (β)0.5LE/ry.7. Determine the bending strength, pb, from Table

4.7 using λ and D/T.8. Calculate the buckling resistance moment, Mb,

via equation 4.18 or 4.19.9. Check Mx ≤ Mb. If unsatisfactory return to (2).

Rigorous method1. Steps (1)–(4) as for conservative method.2. Determine Sx and ry from steel tables; deter-

mine x and u from either steel tables or for UBand UCs assuming x = D/T and u = 0.9.

3. Calculate slenderness ratio, λ = LE/ry.4. Determine ν from Table 4.8 using λ/x and

N = 0.5.5. Calculate equivalent slenderness ratio, λLT =

uνλ√βw.6. Determine pb from Table 4.9 using λLT.7. Calculate Mb via equations 4.22–4.25.8. Obtain the equivalent uniform moment factor,

mLT, from Table 4.10.9. Check Mx ≤ Mb/mLT. If unsatisfactory return to

(2).

4.9 Design of compressionmembers

4.9.1 STRUTSSteel compression members, commonly referred toas stanchions, include struts and columns. A strutis a member subject to direct compression only.A column, on the other hand, refers to memberssubject to a combination of compressive loadingand bending. Although most columns in real struc-tures resist compressive loading and bending, thestrut is a convenient starting point.

Struts (and columns) differ fundamentally in theirbehaviour under axial load depending on whetherthey are slender or stocky. Most real struts andcolumns can neither be regarded as slender norstocky, but as something in between, but let us lookat the behaviour of stocky and slender struts first.

Stocky struts will fail by crushing or squashingof the material. For stocky struts the ‘squash load’,Ps is given by the simple formula

Ps = pyAg (4.27)

wherepy design strength of steelAg gross cross sectional area of the section

Design of compression members

Str

engt

h of

str

ut p

c (N

mm

−2)

300

250

200

150

100

50

050 100 150 200 250 300 350

Slenderness ratio λ

Euler buckling stress pE

Table 24(a)Table 24(b)Table 24(c)Table 24(d)

Yield stress p y = 275 N mm−2

F

F

Fig. 4.23

Slender struts will fail by buckling. For elasticslender struts pinned at each end, the ‘Euler load’,at which a perfect strut buckles elastically is given by

P

EIL

EA rL

EAE 2

g y2

2

g

2= = =

π π πλ

2 2 2

(4.28)

using r

IA

= and λ .=

Lr

If the compressive strength, pc, which is given by

pc = Ps /Ag (for stocky struts) (4.29)

and

pc = PE/Ag (for slender struts) (4.30)

are plotted against λ (Fig. 4.23), the area abovethe two dotted lines represents an impossible situ-ation in respect of these struts. In this area, thestrut has either buckled or squashed. Struts whichfall below the dotted lines are theoretically able towithstand the applied load without either bucklingor squashing. In reality, however, this tends not tobe the case because of a combination of manufac-turing and practical considerations. For example,struts are never completely straight, or are subjectto exactly concentric loading. During manufac-ture, stresses are locked into steel members whicheffectively reduce their load-carrying capacities.As a result of these factors, failure of a strut willnot be completely due to buckling or squashing,but a combination, with partially plastic stressesappearing across the member section. These non-ideal factors or imperfections are found in practice,and laboratory tests have confirmed that in fact thefailure line for real struts lies along a series of linessuch as a,b,c and d in Fig. 4.23. These are derivedfrom the Perry-Robertson equation which includesallowances for the various imperfections.

9780415467193_C04 9/3/09, 1:28 PM177


178

Whichever of the lines a–d is used dependson the shape of section and the axis of buckling.Table 23 of BS 5950, part of which is reproducedas Table 4.12, specifies which of the lines is appro-priate for the shape of section, and Tables 24(a),(b), (c) and (d) enable values of pc to be read off

appropriate to the section used. (Tables 24(b) and(c) of BS 5950 have been reproduced as Tables 4.13and 4.14 respectively.) Alternatively, Appendix Cof BS 5950 gives the actual Perry-Robertson equa-tions which may be used in place of the tables ifconsidered necessary.

Table 4.13 Compressive strength, pc (N/mm2) with λ < 110 for strut curve b(Table 24(b), BS 5950)

λ Steel grade and design strength py (N/mm2)

S275 S355 S460

235 245 255 265 275 315 325 335 345 355 400 410 430 440 460

15 235 245 255 265 275 315 325 335 345 355 399 409 428 438 45720 234 243 253 263 272 310 320 330 339 349 391 401 420 429 44825 229 239 248 258 267 304 314 323 332 342 384 393 411 421 43930 225 234 243 253 262 298 307 316 325 335 375 384 402 411 42935 220 229 238 247 256 291 300 309 318 327 366 374 392 400 41740 216 224 233 241 250 284 293 301 310 318 355 364 380 388 40442 213 222 231 239 248 281 289 298 306 314 351 359 375 383 39944 211 220 228 237 245 278 286 294 302 310 346 354 369 377 39246 209 218 226 234 242 275 283 291 298 306 341 349 364 371 38648 207 215 223 231 239 271 279 287 294 302 336 343 358 365 37950 205 213 221 229 237 267 275 283 290 298 330 337 351 358 37252 203 210 218 226 234 264 271 278 286 293 324 331 344 351 36454 200 208 215 223 230 260 267 274 281 288 318 325 337 344 35656 198 205 213 220 227 256 263 269 276 283 312 318 330 336 34758 195 202 210 217 224 252 258 265 271 278 305 311 322 328 33960 193 200 207 214 221 247 254 260 266 272 298 304 314 320 33062 190 197 204 210 217 243 249 255 261 266 291 296 306 311 32064 187 194 200 207 213 238 244 249 255 261 284 289 298 302 31166 184 191 197 203 210 233 239 244 249 255 276 281 289 294 30168 181 188 194 200 206 228 233 239 244 249 269 273 281 285 29270 178 185 190 196 202 223 228 233 238 242 261 265 272 276 28272 175 181 187 193 198 218 223 227 232 236 254 257 264 267 27374 172 178 183 189 194 213 217 222 226 230 246 249 255 258 26476 169 175 180 185 190 208 212 216 220 223 238 241 247 250 25578 166 171 176 181 186 203 206 210 214 217 231 234 239 241 246

Table 4.12 Strut table selection (based on Table 23, BS 5950)

Type of section Thicknessa Axis of buckling

x–x y–y

Hot-finished structural hollow section 24(a) 24(a)Rolled I-section Up to 40 mm 24(a) 24(b)Rolled H-section Up to 40 mm 24(b)b 24(c)c

Over 40 mm 24(c) 24(d)

Notes. a For thicknesses between 40 and 50 mm the value of pc may be taken as theaverage of the values for thicknesses up to 40 mm and over 40 mm.b Reproduced as Table 4.13.c Reproduced as Table 4.14.

9780415467193_C04 9/3/09, 1:29 PM178

179

80 163 168 172 177 181 197 201 204 208 211 224 226 231 233 23782 160 164 169 173 177 192 196 199 202 205 217 219 223 225 22984 156 161 165 169 173 187 190 193 196 199 210 212 216 218 22186 153 157 161 165 169 182 185 188 190 193 203 205 208 210 21388 150 154 158 161 165 177 180 182 185 187 196 198 201 203 20690 146 150 154 157 161 172 175 177 179 181 190 192 195 196 19992 143 147 150 153 156 167 170 172 174 176 184 185 188 189 19294 140 143 147 150 152 162 165 167 169 171 178 179 182 183 18596 137 140 143 146 148 158 160 162 164 165 172 173 176 177 17998 134 137 139 142 145 153 155 157 159 160 167 168 170 171 173

100 130 133 136 138 141 149 151 152 154 155 161 162 164 165 167102 127 130 132 135 137 145 146 148 149 151 156 157 159 160 162104 124 127 129 131 133 141 142 144 145 146 151 152 154 155 156106 121 124 126 128 130 137 138 139 141 142 147 148 149 150 151108 118 121 123 125 126 133 134 135 137 138 142 143 144 145 147110 115 118 120 121 123 129 130 131 133 134 138 139 140 141 142112 113 115 117 118 120 125 127 128 129 130 134 134 136 136 138114 110 112 114 115 117 122 123 124 125 126 130 130 132 132 133116 107 109 111 112 114 119 120 121 122 122 126 126 128 128 129118 105 106 108 109 111 115 116 117 118 119 122 123 124 124 125120 102 104 105 107 108 112 113 114 115 116 119 119 120 121 122122 100 101 103 104 105 109 110 111 112 112 115 116 117 117 118124 97 99 100 101 102 106 107 108 109 109 112 112 113 114 115126 95 96 98 99 100 103 104 105 106 106 109 109 110 111 111128 93 94 95 96 97 101 101 102 103 103 106 106 107 107 108130 90 92 93 94 95 98 99 99 100 101 103 103 104 105 105135 85 86 87 88 89 92 93 93 94 94 96 97 97 98 98140 80 81 82 83 84 86 87 87 88 88 90 90 91 91 92145 76 77 78 78 79 81 82 82 83 83 84 85 85 86 86150 72 72 73 74 74 76 77 77 78 78 79 80 80 80 81155 68 69 69 70 70 72 72 73 73 73 75 75 75 76 76160 64 65 65 66 66 68 68 69 69 69 70 71 71 71 72165 61 62 62 62 63 64 65 65 65 65 66 67 67 67 68170 58 58 59 59 60 61 61 61 62 62 63 63 63 64 64175 55 55 56 56 57 58 58 58 59 59 60 60 60 60 60180 52 53 53 53 54 55 55 55 56 56 56 57 57 57 57185 50 50 51 51 51 52 52 53 53 53 54 54 54 54 54190 48 48 48 48 49 50 50 50 50 50 51 51 51 51 52195 45 46 46 46 46 47 47 48 48 48 49 49 49 49 49200 43 44 44 44 44 45 45 45 46 46 46 46 47 47 47210 40 40 40 40 41 41 41 41 42 42 42 42 42 43 43220 36 37 37 37 37 38 38 38 38 38 39 39 39 39 39230 34 34 34 34 34 35 35 35 35 35 35 36 36 36 36240 31 31 31 31 32 32 32 32 32 32 33 33 33 33 33250 29 29 29 29 29 30 30 30 30 30 30 30 30 30 30260 27 27 27 27 27 27 28 28 28 28 28 28 28 28 28270 25 25 25 25 25 26 26 26 26 26 26 26 26 26 26280 23 23 23 23 24 24 24 24 24 24 24 24 24 24 24290 22 22 22 22 22 22 22 22 22 22 23 23 23 23 23300 20 20 21 21 21 21 21 21 21 21 21 21 21 21 21310 19 19 19 19 19 20 20 20 20 20 20 20 20 20 –320 18 18 18 18 18 18 18 19 19 19 19 19 19 19 19330 17 17 17 17 17 17 17 17 17 18 18 18 18 18 18340 16 16 16 16 16 16 16 16 17 17 17 17 17 17 17350 15 15 15 15 15 16 16 16 16 16 16 16 16 16 16



S275 S355 S460

235 245 255 265 275 315 325 335 345 355 400 410 430 440 460

9780415467193_C04 9/3/09, 1:29 PM179


180

Table 4.14 Compressive strength, pc (N/mm2) for strut curve c (Table 24(c), BS 5950)


S275 S355 S460

235 245 255 265 275 315 325 335 345 355 400 410 430 440 460

15 235 245 255 265 275 315 325 335 345 355 398 408 427 436 45520 233 242 252 261 271 308 317 326 336 345 387 396 414 424 44225 226 235 245 254 263 299 308 317 326 335 375 384 402 410 42830 220 228 237 246 255 289 298 307 315 324 363 371 388 396 41335 213 221 230 238 247 280 288 296 305 313 349 357 374 382 39740 206 214 222 230 238 270 278 285 293 301 335 343 358 365 38042 203 211 219 227 235 266 273 281 288 296 329 337 351 358 37344 200 208 216 224 231 261 269 276 284 291 323 330 344 351 36546 197 205 213 220 228 257 264 271 279 286 317 324 337 344 35748 195 202 209 217 224 253 260 267 274 280 311 317 330 337 34950 192 199 206 213 220 248 255 262 268 275 304 310 323 329 34152 189 196 203 210 217 244 250 257 263 270 297 303 315 321 33354 186 193 199 206 213 239 245 252 258 264 291 296 308 313 32456 183 189 196 202 209 234 240 246 252 258 284 289 300 305 31558 179 186 192 199 205 229 235 241 247 252 277 282 292 297 30660 176 183 189 195 201 225 230 236 241 247 270 274 284 289 29862 173 179 185 191 197 220 225 230 236 241 262 267 276 280 28964 170 176 182 188 193 215 220 225 230 235 255 260 268 272 28066 167 173 178 184 189 210 215 220 224 229 248 252 260 264 27168 164 169 175 180 185 205 210 214 219 223 241 245 252 256 26270 161 166 171 176 181 200 204 209 213 217 234 238 244 248 25472 157 163 168 172 177 195 199 203 207 211 227 231 237 240 24674 154 159 164 169 173 190 194 198 202 205 220 223 229 232 23876 151 156 160 165 169 185 189 193 196 200 214 217 222 225 23078 148 152 157 161 165 180 184 187 191 194 207 210 215 217 22280 145 149 153 157 161 176 179 182 185 188 201 203 208 210 21582 142 146 150 154 157 171 174 177 180 183 195 197 201 203 20784 139 142 146 150 154 167 169 172 175 178 189 191 195 197 20186 135 139 143 146 150 162 165 168 170 173 183 185 189 190 19488 132 136 139 143 146 158 160 163 165 168 177 179 183 184 18790 129 133 136 139 142 153 156 158 161 163 172 173 177 178 18192 126 130 133 136 139 149 152 154 156 158 166 168 171 173 17594 124 127 130 133 135 145 147 149 151 153 161 163 166 167 17096 121 124 127 129 132 141 143 145 147 149 156 158 160 162 16498 118 121 123 126 129 137 139 141 143 145 151 153 155 157 159

100 115 118 120 123 125 134 135 137 139 140 147 148 151 152 154102 113 115 118 120 122 130 132 133 135 136 143 144 146 147 149104 110 112 115 117 119 126 128 130 131 133 138 139 142 142 144106 107 110 112 114 116 123 125 126 127 129 134 135 137 138 140108 105 107 109 111 113 120 121 123 124 125 130 131 133 134 136110 102 104 106 108 110 116 118 119 120 122 126 127 129 130 132112 100 102 104 106 107 113 115 116 117 118 123 124 125 126 128114 98 100 101 103 105 110 112 113 114 115 119 120 122 123 124116 95 97 99 101 102 108 109 110 111 112 116 117 118 119 120118 93 95 97 98 100 105 106 107 108 109 113 114 115 116 117120 91 93 94 96 97 102 103 104 105 106 110 110 112 112 113

9780415467193_C04 9/3/09, 1:29 PM180

181



S275 S355 S460

235 245 255 265 275 315 325 335 345 355 400 410 430 440 460

122 89 90 92 93 95 99 100 101 102 103 107 107 109 109 110124 87 88 90 91 92 97 98 99 100 100 104 104 106 106 107126 85 86 88 89 90 94 95 96 97 98 101 102 103 103 104128 83 84 86 87 88 92 93 94 95 95 98 99 100 100 101130 81 82 84 85 86 90 91 91 92 93 96 96 97 98 99135 77 78 79 80 81 84 85 86 87 87 90 90 91 92 92140 72 74 75 76 76 79 80 81 81 82 84 85 85 86 87145 69 70 71 71 72 75 76 76 77 77 79 80 80 81 81150 65 66 67 68 68 71 71 72 72 73 75 75 76 76 76155 62 63 63 64 65 67 67 68 68 69 70 71 71 72 72160 59 59 60 61 61 63 64 64 65 65 66 67 67 67 68165 56 56 57 58 58 60 60 61 61 61 63 63 64 64 64170 53 54 54 55 55 57 57 58 58 58 60 60 60 60 61175 51 51 52 52 53 54 54 55 55 55 56 57 57 57 58180 48 49 49 50 50 51 52 52 52 53 54 54 54 54 55185 46 46 47 47 48 49 49 50 50 50 51 51 52 52 52190 44 44 45 45 45 47 47 47 47 48 49 49 49 49 49195 42 42 43 43 43 45 45 45 45 45 46 46 47 47 47200 40 41 41 41 42 43 43 43 43 43 44 44 45 45 45210 37 37 38 38 38 39 39 39 40 40 40 40 41 41 41220 34 34 35 35 35 36 36 36 36 36 37 37 37 37 38230 31 32 32 32 32 33 33 33 33 34 34 34 34 34 35240 29 29 30 30 30 30 31 31 31 31 31 31 32 32 32250 27 27 27 28 28 28 28 28 29 29 29 29 29 29 29260 25 25 26 26 26 26 26 26 27 27 27 27 27 27 27270 23 24 24 24 24 24 25 25 25 25 25 25 25 25 25280 22 22 22 22 22 23 23 23 23 23 23 24 24 24 24290 21 21 21 21 21 21 21 22 22 22 22 22 22 22 22300 19 19 20 20 20 20 20 20 20 20 21 21 21 21 21310 18 18 18 19 19 19 19 19 19 19 19 19 19 19 20320 17 17 17 17 18 18 18 18 18 18 18 18 18 18 18330 16 16 16 16 17 17 17 17 17 17 17 17 17 17 17340 15 15 15 16 16 16 16 16 16 16 16 16 16 16 16350 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15


4.9.2 EFFECTIVE LENGTHAs mentioned in section 4.9.1, the compressivestrength of struts is primarily related to their slen-derness ratio. The slenderness ratio, λ, is given by

λ =

LrE (4.31)

whereLE effective length of the memberr radius of gyration obtained from steel tables.

The concept of effective length was discussed insection 4.8.11.1, in the context of lateral torsionalbuckling, and is similarly applicable to the design ofstruts and columns. The effective length is simplya function of the actual length of the member andthe restraint at the member ends.

The formulae in Appendix C of BS 5950 and thegraph in Fig. 4.23 relate to standard restraint con-ditions in which each end is pinned. In reality eachend of the strut may be free, pinned, partially fixed,

9780415467193_C04 9/3/09, 1:29 PM181


182

LE =

(0.

5) 0

.7L

LE =

0.8

5L

LE =

(0.

7) 0

.85L

LE =

(1.

0) 1

.0L

LE =

(1.

0) 1

.2L

LE =

1.5

L

LE =

(2.

0) 2

.0L

(1) (2) (3) (4) (5) (6) (7)

Fig. 4.24

Table 4.15 Nominal effective length, LE, for a compression member (Table 22, BS 5950)

a) non-sway mode

Restraint (in the plane under consideration) by other parts of the structure LE

Effectively held in position at both ends Effectively restrained in direction at both ends (1) 0.7LPartially restrained in direction at both ends (2) 0.85LRestrained in direction at one end (3) 0.85LNot restrained in direction at either end (4) 1.0L

b) sway mode

One end Other end

Effectively held in position and restrained Not held Effectively restrained in direction (5) 1.2Lin direction in position Partially restrained in direction (6) 1.5L

Not restrained in direction (7) 2.0L

that the design effective lengths are greater thanthe theoretical values where one or both ends of themember are partially or wholly restrained. This isbecause, in practice, it is difficult if not impossibleto guarantee that some rotation of the member willnot take place. Furthermore, the effective lengths arealways less than the actual length of the compres-sion member except when the structure is unbraced.

or fully fixed (rotationally). Also, whether or not thetop of the strut is allowed to move laterally withrespect to the bottom end is important i.e. whetherthe structure is braced or unbraced. Figure 4.24summarises these restraints, and Table 22 of BS5950, reproduced above as Table 4.15 stipulatesconservative assumptions of effective length LE fromwhich the slenderness λ can be calculated. Note

9780415467193_C04 9/3/09, 1:29 PM182

183

Example 4.10 Design of an axially loaded column (BS 5950)A proposed 5 metre long internal column in a ‘rigid’ jointed steel structure is to be loaded concentrically with 1000 kNdead and 1000 kN imposed load (Fig. 4.25 ). Assuming that fixity at the top and bottom of the column gives effectiverotational restraints, design column sections assuming the structure will be (a) braced and (b) unbraced.


Fig. 4.25

1000 kN dead load1000 kN imposed load

Floors

Columns

BRACED COLUMN

Design axial loadingFactored loading, Fc = (1.4 × 1000) + (1.6 × 1000) = 3000 kN

Effective lengthFor the braced case the column is assumed to be effectively held in position at both ends, and restrained in directionat both ends. It will buckle about the weak (y–y) axis. From Table 4.15 therefore, the effective length, LE, is

LE = 0.7L = 0.7 × 5 = 3.5 m

Section selectionThis column design can only really be done by trial and error.Initial trial. Try 254 × 254 × 107 UC:

py = 265 N/mm2 ry = 65.7 mm Ag = 13 700 mm2 b/T = 6.3 d/t = 15.4

λ = LE/ry = 3500/65.7 = 53

From Table 4.12, use Table 24(c) of BS 5950 (i.e. Table 4.14 ), from which pc = 208 N/mm2. UC section is not slendersince b/T < 15ε = 15 × (275/265)0.5 = 15.28 and d/t < 40ε = 40.74 (Table 4.4). From clause 4.7.4 of BS 5950,compression resistance of column, Pc, is

Pc = Agpc = 13700 × 208/103 = 2850 kN < 3000 kN Not OK

Second trial. Try 305 × 305 × 118 UC:

p y = 265 N/mm2 r y = 77.5 mm Ag = 15 000 mm2 b/T = 8.20 d/t = 20.7

λ = LE /ry = 3500/77.5 = 45

Then from Table 24(c) of BS 5950 (i.e. Table 4.14 ), pc = 222 N/mm2. UC section is not slender then

Pc = Agpc = 15000 × 222/103 = 3330 kN > 3000 kN OK

9780415467193_C04 9/3/09, 1:30 PM183


184

4.9.3 COLUMNS WITH BENDING MOMENTSAs noted earlier, most columns in steel structuresare subject to both axial load and bending. Ac-cording to clause 4.8.3.1 of BS 5950, such membersshould be checked (for yielding or local buckling)at the points of greatest bending moment and axialload, which usually occur at the member ends. Inaddition, the buckling resistance of the member asa whole should be checked.

4.9.3.1 Cross-section capacity checkThe purpose of this check is to ensure thatnowhere across the section does the steel stressexceed yield. Generally, except for class 4 slendercross-sections, clause 4.8.3.2 states that the follow-ing relationship should be satisfied:

FA p

MM

MM

c

g y

x

cx

y

cy

+ + ≤ 1 (4.32)

whereFc axial compression loadAg gross cross-sectional area of sectionpy design strength of steelMx applied major axis momentMcx moment capacity about the major axis in

the absence of axial loadMy applied minor axis momentMcy moment capacity about the minor axis in

the absence of axial load

When the section is slender the expression inclause 4.8.3.2 (c) should be used. Note thatparagraph (b) of this clause gives an alternativeexpression for calculating the (local) capacity ofcompression members of class 1 plastic or class 2compact cross-section, which yields a more exactestimate of member strength.

4.9.3.2 Buckling resistance checkBuckling due to imposed axial load, lateral tor-sional buckling due to imposed moment, or a com-bination of buckling and lateral torsional bucklingare additional possible modes of failure in mostpractical columns in steel structures.

Clause 4.8.3.3.1 of BS 5950 gives a simplifiedapproach for calculating the buckling resistance ofcolumns which involves checking that the follow-ing relationships are both satisfied:

FP

m Mp Z

m Mp Z

c

c

x x

y x

y y

y y

+ + ≤ 1 (4.33)

FP

m MM

m Mp Z

c

cy

LT LT

b

y y

y y

+ + ≤ 1 (4.34)

wherePc smaller of Pcx and Pcy

Pcx compression resistance of memberconsidering buckling about the major axis

Pcy compression resistance of memberconsidering buckling about the minor axis

mx equivalent uniform moment factor formajor axis flexural buckling obtainedfrom Table 26 of BS 5950, reproducedas Table 4.16

my equivalent uniform moment factor for minoraxis flexural buckling obtained from Table26 of BS 5950, reproduced as Table 4.16

mLT equivalent uniform moment factor for lateraltorsional buckling obtained from Table 18of BS 5950, reproduced as Table 4.10

MLT maximum major axis moment in thesegment length Lx governing Pcx

Mb buckling resistance momentZx elastic modulus about the major axisZy elastic modulus about the minor axis and

the other symbols are as above.

UNBRACED COLUMNFor the unbraced case, LE = 1.2L = 6.0 metres from Table 4.15, and the most economic member would appear to be305 × 305 × 158 UC:

py = 265 N/mm2 ry = 78.9 mm Ag = 20 100 mm2 b/T = 6.21 d/t = 15.7

λ = LE/r y = 6000/78.9 = 76

Then from Table 4.14, pc = 165 N/mm2. Section is not slender

Pc = Agpc = 20 100 × 165/103 = 3317 kN > 3000 kN OK

Hence, it can immediately be seen that for a given axial load, a bigger steel section will be required if the column isunbraced.


9780415467193_C04 9/3/09, 1:31 PM184

185

Table 4.16 Equivalent uniform moment factor m for flexural buckling(Table 26, BS 5950)

1.000.960.920.880.840.800.760.720.680.640.600.580.560.540.520.500.480.460.440.420.40

Segments with end moments only

β positive

x

x

M

L

βM

x x

x xβM

M

L

Segments between intermediate lateral restraints

Specific cases

x xL

x x

Lx x

Lx x

L

m = 0.90 m = 0.95 m = 0.95 m = 0.80

β m

1.00.90.80.70.60.50.40.30.20.10.0

−0.1−0.2−0.3−0.4−0.5−0.6−0.7−0.8−0.9−1.0

β negative


Example 4.11 Column resisting an axial load and bending (BS 5950)Select a suitable column section in grade S275 steel to support a factored axial concentric load of 2000 kN andfactored bending moments of 100 kN m about the major axis, and 20 kN m about the minor axis (Fig. 4.26), appliedat both ends of the column. The column is 10 m long and is fully fixed against rotation at top and bottom, and thefloors it supports are braced against sway.

INITIAL SECTION SELECTION305 × 305 × 118 UC:

py = 265 N/mm2, plastic Sx = 1950 cm3 Z x = 1760 cm3

Ag = 150 cm2 Sy = 892 cm3 Z y = 587 cm3

t = 11.9 mm x = D/T = 314.5/18.7 u = 0.9d = 246.6 mm = 16.8ry = 7.75 cm

Note: In this case the classification procedure is slightly different in respect of web classification. From Table 11 ofBS 5950 r1 = Fc/dtpy but −1 < r1 ≤ 1

= 2 × 106/246.6 × 11.9 × 265 = 2.87

Therefore, r1 = 1. Limiting d /t = 80ε/1 + r1 =

80

11 +ε = 40(276/265)1/2 = 40.75

9780415467193_C04 9/3/09, 1:31 PM185


186


2000 kN

20 kN m100 kN m

10 m

Fig. 4.26

Actual d /t = 20.7, hence web is plastic.b/T = 8.20 < 9ε = 9(275/265)1/2 = 9.17, hence flange is plastic.

Mcx = pySx = 265 × 1 950 × 10−3 = 516.75 kN m

Mcy = pySy = 265 × 892 × 10−3 = 236.38 kN m

CROSS-SECTION CAPACITY CHECK

Substituting into

FA p

MM

MM

c

g y

x

cx

y

cy + + gives

2000 10150 10

3

2

×× ×

+ + = + + = <

.

.

. . . . 265

100516 75

20236 38

0 503 0 193 0 085 0 781 1 OK

BUCKLING RESISTANCE CHECKIn-plane bucklingFrom Table 4.15, effective length LE = 0.7L = 7 m

λ = L E/ry = 7000/77.5 = 90

From Table 4.12, for buckling about the x–x axis and y–y axis use, respectively, Table 24(b) of BS 5950 (Table 4.13)from which pcx = 157 N/mm2, and Table 24(c) of BS 5950 (Table 4.14) from which pcy = 139 N/mm2. Then

Pcx = Ag pcx = 150 × 102 × 157 × 10−3 = 2355 kN

Pcy = Ag pcy = 150 × 102 × 139 × 10−3 = 2085 kN = Pc

Ratio of end moments about both x–x and y–y axes, β = 1. Hence from Table 4.16, m x = m y = 1

Substituting into

FP

m Mp Z

m Mp Z

c

c

x x

y x

y y

y y

gives + +

20002085 10 103 3

+×

× ×+

×× ×− −

1 100265 1760

1 20265 587

= 0.96 + 0.21 + 0.13 = 1.30 > 1 Not OK

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187


Example 4.11 continuedLateral torsional buckling

( ) ( . )

. . .βw

E

y

0 5 0 51 0700077 5

90Lr = = and D/T = 16.8

From Table 4.7, pb = 196 N/mm2

Mb = pbSx = 196 × 1950 × 10−3 = 382.2 kN m

Ratio of end moments about both major axes, β = 1. Hence from Table 4.10, m LT = 1

Substituting into

FP

m MM

m Mp Z

c

cy

LT LT

b

y y

y y

gives + +

20002085 382 2 10 3

.

+×

+×

× × −

1 100 1 20265 587

= 0.96 + 0.26 + 0.13 = 1.35 > 1 Not OK

Hence, a bigger section should be selected.

SECOND SECTION SELECTIONTry 356 × 368 × 177 UC:

py = 265 N/mm2, plastic Sx = 3460 cm3 Z x = 3100 cm3

Ag = 226 cm2 Sy = 1670 cm3 Zy = 1100 cm3

ry = 9.52 cm3 x = D/T = 368.3/23.8 u = 0.9= 15.5

Mcx = pySx = 265 × 3460 × 10−3 = 916.9 kN m

Mcy = py Sy = 265 × 1670 × 10−3 = 442.55 kN m

CROSS-SECTION CAPACITY CHECK

Again using

FA p

MM

MM

c

g y

x

cx

y

cy

gives + +

2000 10226 10

3

2

×× ×

+ +

.

.265

100916 9

20442 55

= 0.33 + 0.11 + 0.05 = 0.49 < 1 OK

BUCKLING RESISTANCE CHECKIn-plane bucklingFrom Table 4.15, effective length LE = 0.7L = 7 m

λ = L E /ry = 7000/95.2 = 73.5

From Table 4.12, for buckling about the x–x axis and y–y axis use, respectively, Table 24(b) of BS 5950 (Table 4.13)from which pcx = 190 N/mm2, and Table 24(c) of BS 5950 (Table 4.14 ) from which pcy = 169 N/mm2. Then

Pcx = Ag pcx = 226 × 102 × 190 × 10−3 = 4294 kN

Pcy = Ag pcy = 226 × 102 × 169 × 10−3 = 3819.4 kN = Pc

Ratio of end moments about both x–x and y–y axes, β = 1. Hence from Table 4.16, mx = my = 1

Substituting into

FP

m Mp Z

m Mp Z

c

c

x x

y x

y y

y y

gives + +

20003819 4 10 103 3.

+×

× ×+

×× ×− −

1 100265 3100

1 20265 1100

= 0.52 + 0.12 + 0.07 = 0.71 < 1 OK

9780415467193_C04 9/3/09, 1:32 PM187


188

Fig. 4.27 Load eccentricity for columns in simple construction.

Beam R

Beam S

Section L–L

(a)

x ex

D/2 100

100

t/2

x Fx

y

Fy

Load frombeam S D

Load frombeam R

A

L

B

L

Loading at column edgedue to beam on cap plate

(c)(b)

Example 4.11 continuedLateral torsional buckling

( ) ( . )

. .. .βw

E

y

0 5 0 51 0700095 2

73 5Lr

= = and D/T = 15.5


Mb = pbSx = 220 × 3460 × 10−3 = 761.2 kN m

Ratio of end moments about both major axes, β = 1. Hence from Table 4.10, m LT = 1

Substituting into

FP

m MM

m Mp Z

c

cy

LT LT

b

y y

y y

gives + + ≤ 1

20003819 4 761 2 10 3.

.

+×

+×

× × −1 100 1 20

265 1100 = 0.52 + 0.13 + 0.07 = 0.72 < 1 OK

Hence a 356 × 368 × 177 UC section is satisfactory.

Clause 4.8.3.3.2 of BS 5950 gives a more exactapproach, but as in practice most designers tend touse the simplified approach, the more exact methodis not discussed here.

4.9.4 COLUMN DESIGN IN ‘SIMPLE’CONSTRUCTION

At first sight it would appear that columns inso-called ‘simple construction’ are not subject tomoments, as the beams are all joined at connec-tions which allow no moment to develop. In fact,

in most cases there is a bending moment due tothe eccentricity of the shear load from the beam.This is summarised in Clause 4.7.7 of BS 5950and illustrated in Fig. 4.27. Note that where abeam sits on a column cap plate, for example at A(Fig. 4.27(c)), it can be assumed that the reactionfrom the beam acts at the face of the column. How-ever, where the beam is connected to a columnby means of a ‘simple’ connection, e.g. using webcleats, the reaction from the beam can be assumedto act 100 mm from the column (web or flange)face as illustrated in Fig. 4.27(b).

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189

whereFc axial compressive loadPc = Agpc – for all classes except class 4 (clause

4.7.4 of BS 5950) – in which Ag is the grosscross-sectional area of the section and pc isthe compressive strength, see section 4.9.1

Mx nominal major axis momentMy nominal minor axis momentMbs buckling resistance moment for ‘simple’

columnspy design strength of steelZy elastic modulus about the minor axis.

Note that this expression is similar to that usedfor checking the buckling resistance of columns incontinuous structures (equation 4.34), but with allequivalent moment factors taken as 1.0. For I- andH-sections Mbs = Mb determined as discussed insection 4.8.11.3 but using the equivalent slender-ness of the column, λLT, given by

λLT = 0.5L /ry (4.36)

whereL is the distance between levels at which the

column is laterally restrained in both directionsry is the radius of gyration about the minor axis.

Fig. 4.29


Ultimatereaction fromBeam B – 75 kN

Ultimatereaction fromBeam A – 200 kN

Plan on capL = 7 m

F

Ultimate load dueto self-weight 5 kN

LE = 0.85L

Effectively held in positionbut not restrained indirection, i.e. pinned

Effectively held in positionand direction, i.e. fixed

Fig. 4.28 Column supporting a roof truss.

F

Roof truss

ColumnCap plate

Example 4.12 Design of a steel column in ‘simple’ construction (BS 5950)Select a suitable column section in S275 steel to support the ultimate loads from beams A and B shown in Fig. 4.29.Assume the column is 7 m long and is effectively held in position at both ends but only restrained in direction at thebottom.

When a roof truss is supported on a column capplate (Fig. 4.28), and the connection is unable todevelop significant moments, it can be assumedthat the load from the truss is transmitted concen-trically to the column.

Columns in simple construction will not needto be checked for local capacity but it will stillbe necessary to check for buckling, which involvessatisfying the following relationship:

FP

MM

Mp Z

c

c

x

bs

y

y y

1 + + ≤ (4.35)

9780415467193_C04 9/3/09, 1:33 PM189


190

SECTION SELECTIONThis can only really be done by trial and error. Therefore, try a: 203 × 203 × 52 UC: Sx = 568 cm3, plastic.

DESIGN LOADING AND MOMENTSUltimate reaction from beam A, RA = 200 kN; ultimate reaction from beam B, RB = 75 kN; assume self-weight ofcolumn = 5 kN. Ultimate axial load, F, is

F = RA + RB + self-weight of column

= 200 + 75 + 5 = 280 kN

Load eccentricity for beam A,

ex = D/2 + 100 = 206.2/2 + 100 = 203.1 mm

Load eccentricity for beam B,

e y = t /2 + 100 = 8/2 + 100 = 104 mm

Moment due to beam A,

M x = RAex = 200 × 103 × 203.1 = 40.62 × 106 N mm

Moment due to beam B,

M y = RBey = 75 × 103 × 104 = 7.8 × 106 N mm

EFFECTIVE LENGTHFrom Table 4.15, effective length coefficient = 0.85. Hence, effective length is

LE = 0.85L = 0.85 × 7000 = 5950 mm

BENDING STRENGTHFrom Table 4.12, relevant compressive strength values for buckling about the x–x axis are obtained from Table 24(b)(Table 4.13) and from Table 24(c) (Table 4.14 ) for bending about the y–y axis.

λ x = LE/rx = 5950/89 = 66.8

From Table 4.13, pc = 208 N/mm2.

λ y = LE/ry = 5950/51.6 = 115.3

From Table 4.14, pc = 103 N/mm2. Hence critical compressive strength of column is 103 N/mm2.

BUCKLING RESISTANCEλ LT = 0.5L /ry = 0.5 × 7000/51.6 = 67.8

From Table 4.9, p b = 193 N/mm2. Buckling resistance moment capacity of column, Mbs, is given by

Mbs = Mb = pbSx = 193 × 568 × 103 = 109.6 × 106 N mm

Hence for stability,

FP

MM

Mp Z

c

c

x

bs

y

y y

1 + + ≤

280 1066.4 10

40.6 10109.6 10

7.8 10275 174 10

3

2

6

6

6

3

×× ×

+××

+×

× ×

103 = 0.41 + 0.37 + 0.16 = 0.94 < 1

Therefore, the 203 × 203 × 52 UC section is suitable.


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191

4.9.5 SUMMARY OF DESIGN PROCEDURES FORCOMPRESSION MEMBERS

Axially loaded members1. Determine ultimate axial load Fc.2. Select trial section and check it is non-slender.3. Determine rx, ry and Ag from steel tables.4. Determine effective lengths, LEX and LEY, using

Table 4.15.5. Calculate slenderness ratios, λEX (= LEX/rx) and

λEY (= LEY/ry).6. Select suitable strut curves from Table 4.12.7. Determine compressive strength, pc, using Table

4.13, 4.14 or similar.8. Calculate compression resistance of member,

Pc = Ag pc.9. Check Fc ≤ Pc. If unsatisfactory return to 2.

Members subject to axial load and bending1. Determine ultimate axial load, Fc, and bend-

ing moments, Mx and My.2. Select and classify trial section.3. Calculate moment capacities of section, Mcx

and Mcy. If either Mx > Mcx or My > Mcy returnto 2.

4. Check cross-section capacity of section viaequation 4.32. If unsatisfactory return to 2.

5. Determine effective lengths, LEX and LEY, usingTable 4.15.

6. Calculate slenderness ratios, λEX (= LEX/rx) andλEY (= LEY /ry).

7. Select suitable strut curves from Table 4.12.8. Determine the major and minor axes com-

pressive strengths, pcx and pcy, using Table 4.13,4.14 or similar.

9. Calculate compressive resistances, Pcx (= pcx Ag)and Pcy (= pcy Ag).

10. Evaluate buckling resistance of section, Mb.11. Determine equivalent uniform moment factors

for flexural buckling, mx and my, using Table4.16.

12. Check buckling resistance of member usingequations (4.33) and (4.34). If unsatisfactoryreturn to 2.

Compression members in simple construction1. Determine ultimate axial load, Fc, and bend-

ing moments, Mx and My.2. Select and classify trial section.3. Determine effective lengths, LEX and LEY, using

Table 4.15.4. Calculate slenderness ratios, λEX (= LEX/rx) and

λEY (= LEY/ry).


6. Select suitable strut curves from Table 4.12.7. Determine compressive strength, pc, using Table

4.13, 4.14 or similar.8. Calculate compression resistance, Pc = Ag pc. If

Pc < Fc return to 2.9. Calculate effective slenderness ratio, λLT =

0.5L/ry.10. Calculate buckling resistance of section, Mbs =

Mb = pbSx.11. Check buckling resistance of member using

equation (4.35). If unsatisfactory return to (2).

4.9.6 DESIGN OF CASED COLUMNSAs discussed in section 4.5, steel columns are some-times cased in concrete for fire protection. How-ever, the concrete also increases the strength of thesection, a fact which can be used to advantage indesign provided that the conditions stated in Clause4.14.1 of BS 5950 are met. Some of these condi-tions are illustrated in Fig. 4.30.

BS 5950 gives guidance on the design of UCsections encased in concrete for the following load-ing conditions which are discussed below.

(i) axially loaded columns(ii) columns subject to axial load and bending.

Longitudinalbars

B

50 min75 max

50 min75 max

Ddc

bc

Characteristicstrength of concrete � 20 N/mm2

Reinforcement

Reinforcement: steel fabric type D98(BS 4483) or � 5 mm diameter longitudinalbars and links at a maximum spacing of 200 mm

= =

==

Fig. 4.30 Cased UC section.

9780415467193_C04 9/3/09, 1:35 PM191


192

select selection

calculate effective length*1, LE

calculate radii of gyration of cased section*2, r y and r x

calculate compression resistance of column*3, Pc

Fig. 4.31 Design procedure for axially loaded cased columns.

Notes to Fig. 4.311 The effective length, LE, should not exceed the

least of:(i) 40bc

(ii)

100bd

c2

c(iii) 250r

where bc and dc are as indicated in Fig. 4.30and r is the minimum radius of gyration of theuncased UC section i.e. ry.

2 The radius of gyration of the cased sectionabout the y–y axis, ry, is taken as 0.2bc but notmore than 0.2(B + 150) mm and not less thanthat of the steel section alone.

The radius of gyration of the cased sectionabout the x–x axis, rx, is taken as that of thesteel section alone.

3 The compression resistance of the cased section,Pc, is given by

P A

f Ap

pc gcu c

yc

.= +

0 45

(4.37)

However, this should not be greater than theshort strut capacity of the section, Pcs, which isgiven by:

P A

f Ap

pcs gcu c

yy

.= +

0 25

(4.38)

whereAc is the gross sectional area of the concrete

but neglecting any casing in excess of75 mm from the overall dimensions ofthe UC section or any applied finish

Ag is the gross sectional area of the UCsection

fcu is the characteristic strength of theconcrete which should not be greater than40 N/mm2

pc is the compressive strength of the UCsection determined as discussed foruncased columns (section 4.9.1), but usingrx and ry for the cased section (see note 2)and taking py ≤ 355 N/mm2

py design strength of the UC section whichshould not exceed 355 N/mm2.

4.9.6.1 Axially loaded columnsThe design procedure for this case is shown inFig. 4.31.

4.9.6.2 Cased columns subject to axial loadand moment

The design procedure here is similar to that whenthe column is axially loaded but also involves check-ing the member’s cross-section capacity and buck-ling resistance using the following relationships:

1. Cross-section capacity check

FP

MM

MM

c

cs

x

cx

y

cy

+ + ≤ 1 (4.39)

whereFc axial compression loadPcs short strut capacity (equation 4.38)Mx applied moment about major axisMcx major axis moment capacity of steel

sectionMy applied moment about minor axisMcy minor axis moment capacity of steel

section2. Buckling resistance check

Major axis

FP

m Mp Z

m Mp Z

c

c

x x

y x

y y

y y

+ + ≤ 1 (4.40)

Minor axis

F

P

m M

M

m M

p Zc

cy

LT x

b

y y

y y

+ + ≤ 1 (4.41)

whereFc maximum compressive axial forcePc smaller of Pcx and Pcy (equation 4.37)Pcx compression resistance of member

considering buckling about the majoraxis

Pcy compression resistance of memberconsidering buckling about theminor axis

9780415467193_C04 9/3/09, 1:36 PM192

193


Example 4.13 Encased steel column resisting an axial load (BS 5950)Calculate the compression resistance of a 305 × 305 × 118 kg/m UC column if it is encased in concrete of com-pressive strength 20 N/mm2 in the manner shown below. Assume that the effective length of the column about bothaxes is 3.5 m.

59B = 306.8

59305 × 305 × 118 kg m–1 UC

55

D = 314.5

55

bc = 425

dc = 425

PROPERTIES OF UC SECTIONArea of UC section (Ag) = 15000 mm2 (Appendix B)Radius of gyration (rx) = 136 mmRadius of gyration (ry) = 77.5 mmDesign strength (py) = 265 N/mm2 (since T = 18.7 mm)Effective length (LE) = 3.5 m

EFFECTIVE LENGTHCheck that the effective length of column (= 3500 mm) does not exceed the least of:

(i) 40bc = 40 × 425 = 17 000 mm

(ii)

100 100 425425

42 5002b

dc2

c

mm

=×

=

(iii) 250ry = 250 × 77.5 = 19 375 mm OK

RADII OF GYRATION FOR THE CASED SECTIONFor the cased section r x is the same as for UC section = 136 mmFor the cased section r y = 0.2bc = 0.2 × 425 = 85 mm � 0.2(B + 150) = 0.2(306.8 + 150) = 91.36 mm but not lessthan that for the uncased section (= 77.5 mm)Hence r y = 85 mm and r x = 136 mm

COMPRESSION RESISTANCESlenderness ratio

λ x

E

x

.= = =Lr

3500136

25 7

λ y

E

y

.= = =Lr

350085

41 2

Compressive strengthFrom Table 4.12, relevant compressive strength values for buckling about the x–x axis are obtained from Table 24(b)of BS 5950 (Table 4.13) and from Table 24(c) of BS 5950 (Table 4.14) for bending about the y–y axis.

9780415467193_C04 9/3/09, 1:36 PM193


194

mx, m y equivalent uniform moment factorsfor major axis and minor axisbuckling respectively obtained fromTable 26 of BS 5950, reproduced asTable 4.16

mLT equivalent uniform moment factorfor lateral torsional bucklingobtained from Table 18 of BS 5950,reproduced as Table 4.10

Mb buckling resistance moment of thecased column = Sxpb ≤ 1.5Mb for theuncased section. To determine pb, ry

should be taken as the greater of ry

of the uncased section or0.2(B + 100) mm (Fig. 4.30).

Mx, My maximum moment about the majorand minor axes respectively

Zx, Zy elastic modulus about the major andminor axes respectively.

Example 4.13 continuedFor λ x = 25.7 and py = 265 N/mm2 compressive strength, pc = 257 N/mm2 (Table 4.13). For λ y = 41.2 and py =

265 N/mm2 compressive strength, pc = 228 N/mm2 (Table 4.14).Hence, pc is equal to 228 N/mm2.

Compression resistanceAg = 15 000 mm2

Ac = dcbc = 425 × 425 = 180 625 mm2

py = 265 N/mm2 (since T = 18.7 mm)

pc = 228 N/mm2

fcu = 20 N/mm2

Compression resistance of encased column, Pc, is given by

P A

f Ap

pc gcu c

yc

.= +

0 45

Pc 15000

20 180625265

.

= +× ×

0 45228 = 4.81 × 106 N = 4810 kN

which should not be greater than the short strut capacity, P cs, given by

P A

f Ap

pcs gcu c

yy

.= +

0 25

15000 20 180625

265

.= +

× ×

0 25265 = 4.878 × 106 N = 4878 kN OK

Hence the compression resistance of the encased column is 4878 kN. Comparing this with the compression resistanceof the uncased column (Example 4.10) shows that the load capacity of the column has been increased from 3300 kNto 4878 kN, which represents an increase of approximately 45%.

9780415467193_C04 9/3/09, 1:37 PM194

195


PROPERTIES OF UC SECTION

Area of UC section, Ag = 15 000 mm2

Radius of gyration about x–x axis, rx = 136 mmRadius of gyration about y–y axis, ry = 77.5 mmElastic modulus about x–x axis, Zx = 1760 × 103 mm3

Elastic modulus about y–y axis, Z y = 587 × 103 mm3

Plastic modulus about x–x axis, S x = 1950 × 103 mm3

Design strength, py = 265 N/mm2 (since T = 18.7 mm)Effective length, LE = 7 m

LOCAL CAPACITY

Axial load, Fc = 2000 kNApplied moment about x–x axis, M x = 100 kN mApplied moment about y–y axis, M y = 20 kN m

Short strut capacity, Pcs, is given by

P A

f Ap

pcs gcu c

yy

.= +

0 25

2

15000 20 425

265

.= +

× ×

0 25265 = 4.878 × 106 N = 4878 kN

Moment capacity of column about the x–x axis, Mcx, is given by

Mcx = pyZx = 265 × 1760 × 103 = 466.4 × 106 N mm = 466.4 kN m

59B = 306.8

59305 × 305 × 118 kg m–1 UC

55

D = 314.5

55

bc = 425

dc = 425

Example 4.14 Encased steel column resisting an axial load and bending(BS 5950)

In Example 4.11 it was found that a 305 × 305 × 118 kg/m UC column was incapable of resisting the design load andmoments below:

Design axial load = 2000 kNDesign moment about x–x axis = 100 kN mDesign moment about y–y axis = 20 kN m

Assuming that the same column is now encased in concrete as show below, determine its suitability. The effectivelength of the column about both axes is 7 m.

9780415467193_C04 11/3/09, 11:08 AM195


196

Example 4.14 continuedMoment capacity of column about the y–y axis, Mcy, is given by

Mcy = pyZy = 265 × 587 × 103 = 155.6 × 106 N mm = 155.6 kN m

FP

MM

MM

c

cs

x

cx

y

cy

.

.

+ + = + +20004878

100466 4

20155 6

= 0.41 + 0.21 + 0.13 = 0.75 < 1

Hence, the local capacity of the section is satisfactory.

BUCKLING RESISTANCERadii of gyration for cased sectionFor the cased section rx is the same as for the UC section = 136 mmFor the cased section ry = 0.2bc = 0.2 × 425 = 85 mm � 0.2(B + 150) = 0.2(306.8 + 150) = 91.36 mm but not lessthan that for the uncased section (= 77.5 mm)Hence r y = 85 mm and rx = 136 mm.

Slenderness ratio

λx

E

x

= = = .Lr

7000136

51 5 λy

E

y

= = = .Lr

700085

82 4

Compressive strengthFor λ x = 51.5 and py = 265 N/mm2 compressive strength, pc = 226 N/mm2 (Table 4.13). For λ y = 82.4 and py =265 N/mm2 compressive strength, pc = 153 N/mm2 (Table 4.14).Hence, pc is equal to 153 N/mm2.

Compression resistanceAg = 15 000 mm2

Ac = dcbc = 425 × 425 = 180 625 mm2

p y = 265 N/mm2 (since T = 18.7 mm)pc = 153 N/mm2

fcu = 20 N/mm2

Compression resistance of encased column, P c, is given by

P A

f Ap

pc gcu c

yc

.= +

0 45

15000 20 180625

265

.= +

× ×

0 45153 = 3.233 × 106 N = 3233 kN

which is not greater than the short strut capacity, Pcs = 4878 kN (see above) OK

Buckling resistanceFor the uncased section,

λy

E

y

. = = =

Lr

700077 5

90

λ y

x

. / . .= =

90314 5 18 7

5 4 ⇒ ν = 0.79 (Table 4.8)

λ LT = uνλ y βw = 0.851 × 0.79 × 90 1 = 61


Mb = Sx pb = 1950 × 103 × 205 × 10−6 = 400 kN m

9780415467193_C04 11/3/09, 11:09 AM196

197


Fig. 4.32

2c + T2c + t

Effective bearing area

4.9.7 DESIGN OF COLUMN BASEPLATESClause 4.13 gives guidance on the design of con-centrically loaded column slab baseplates, whichcovers most practical design situations. The planarea of the baseplate is calculated by assuming

a) the nominal bearing pressure between thebaseplate and support is uniform and

b) the applied load acts over a portion of thebaseplate known as the effective area, the ex-tent of which for UB and UCs is as indicatedon Fig. 4.32.

For concrete foundations the bearing strengthmay be taken as 0.6 times the characteristic cubestrength of the concrete base or the bedding mater-ial (i.e. 0.6fcu), whichever is the lesser. The effectivearea of the baseplate, Abe, is then obtained from

Abe

axial loadbearing strength

= (4.42)

In determining the overall plan area of the plateallowance should be made for the presence of hold-ing bolts.

For the cased section,

λy

E

y

.= = =Lr

700085

82 4

λy

x

.. / .

.= =82 4

314 5 18 74 9 ⇒ ν = 0.82 (Table 4.8)

λ LT = uνλ y βw = 0.851 × 0.82 × 82.4 1 = 57.4


Mb = Sx pb = 1950 × 103 × 213 × 10−6 = 415 kN m

Hence, Mb (for cased UC section = 415 kN m) < 1.5Mb (for uncased UC section = 1.5 × 400 = 600 kN m).

Checking buckling resistance

FP

m Mp Z

m Mp Z

c

c

x x

y x

y y

y y

+ +

20003233

1 100265 1760 10

1 20265 587 103 3

+

×× ×

+×

× ×− − = 0.62 + 0.21 + 0.13 = 0.96 < 1 OK

F

P

m M

M

m M

p Zc

cy

LT x

b

y y

y y

+ +

20003233

1 100415

1 20265 587 10 3

+×

+×

× × − = 0.62 + 0.24 + 0.13 = 0.99 < 1 OK

Hence, the section is now just adequate to resist the design axial load of 2000 kN and design moments about thex–x and y–y axes of 100 kN m and 20 kN m respectively.


9780415467193_C04 9/3/09, 1:38 PM197


198

AREA OF BASEPLATEEffective area

A be

2axial loadbearing strength

mm

. . ≥ =

××

= ×3000 10

0 6 301 666 10

35

Actual areaAbe = (B + 2c)(D + 2c) − 2{(D − 2[T + c])([B + 2c] − [t + 2c])}

1.666 × 105 = (306.8 + 2c)(314.5 + 2c) − 2{(314.5 − 2[18.7 + c])(306.8 − 11.9)} ⇒ c = 84.6 mm

Minimum length of baseplate = D + 2c = 314.5 + 2 × 84.6 = 483.7 mmMinimum width of baseplate = B + 2c = 306.4 + 2 × 84.6 = 476 mmProvide 500 × 500 mm baseplate in grade S275 steel.

BASEPLATE THICKNESSAssuming a baseplate thickness of less than 40 mm the design strength pyp = 265 N/mm2. The actual baseplatethickness, tp, is

tp = c[3ω /pyp]0.5 = 84.6[3 × (0.6 × 30)/265]0.5 = 34.9 mm

Hence, a 500 mm × 500 mm × 35 mm thick baseplate in grade S275 steel should be suitable.

The required minimum baseplate thickness, tp,is given by

tp = c[3ω/pyp]0.5 (4.43)

wherec is the largest perpendicular distance from

the edge of the effective portion of the

baseplate to the face of the columncross-section (Fig. 4.32)

ω pressure on the underside of the plateassuming a uniform distribution throughoutthe effective portion, but ≤ 0.6fcu

pyp design strength of the baseplate which maybe taken from Table 4.3

Example 4.15 Design of a steel column baseplate (BS 5950)Design a baseplate for the axially loaded column shown below assuming it is supported on concrete of compressioncharacteristic strength 30 N/mm2.

Axial load = 3000 kN

B = 306.8 mm

D = 314.5 mm

tp

305 × 305 × 118 kg m–1 UCT = 18.7 mm

b

b

a a

9780415467193_C04 9/3/09, 1:39 PM198

199

Floor systems for steel framed structures

4.10 Floor systems for steelframed structures

In temporary steel framed structures such as carparks and Bailey bridges the floor deck can beformed from steel plates. In more permanent steelframed structures the floors generally comprise:

• precast, prestressed concrete slabs• in-situ reinforced concrete slabs• composite metal deck floors.

Precast floors are normally manufactured usingprestressed hollow core planks, which can easilyspan up to 6– 8 m (Fig. 4.33(a)). The top surfacecan be finished with a levelling screed or, if a com-posite floor is required, with an in-situ concretestructural topping. This type of floor slab offers anumber of advantages over other flooring systemsincluding:

• elimination of shuttering and propping• reduced floor depth by supporting the precast

units on shelf angles Fig. 4.33(b)• rapid construction since curing or strength devel-

opment of the concrete is unnecessary.

Precast floors are generally designed to act non-compositely with the supporting steel floor beams.Nevertheless, over recent years, hollow core planksthat act compositely with the floor beams have beendeveloped and are slowly beginning to be speci-fied. A major drawback of precast concrete slabs,which restricts their use in many congested citycentre developments, is that the precast units areheavy and cranage may prove difficult. In suchlocations, in-situ concrete slabs have invariablybeen found to be more practical.

In-situ reinforced concrete floor slabs can beformed using conventional removable shutteringand are normally designed to act compositely withthe steel floor beams. Composite action is achievedby welding steel studs to the top flange of the steelbeams and embedding the studs in the concretewhen cast (Fig. 4.34). The studs prevent slippageand also enable shear stresses to be transferred be-tween the slab and supporting beams. This increasesboth the strength and stiffnesses of the beams,thereby allowing significant reductions in construc-tion depth and weight of steel beams to be achieved(see Example 4.16 ). Composite construction notonly reduces frame loadings but also results in

Fig. 4.34 In-situ reinforced concrete slab.

Steel studswelded to topflange of steelbeam

Reinforcedconcrete slab

Fig. 4.33 Precast concrete floor: (a) hollow core plank-section (b) precast concrete plank supported on shelf angles.

(a)

(b)

9780415467193_C04 9/3/09, 1:39 PM199


200

Beam A Beam B600300

300

250 250

300

300

8 m

Example 4.16 Advantages of composite construction (BS 5950)Two simply supported, solid steel beams 250 mm wide and 600 mm deep are required to span 8 m. Both beams aremanufactured using two smaller beams, each 250 mm wide and 300 mm deep, positioned one above the other. Inbeam A the two smaller beams are not connected but act independently whereas in beam B they are fully joined andact together as a combined section.

(a) Assuming the permissible strength of steel is 165 N/mm2, determine the maximum uniformly distributed load thatBeam A and Beam B can support.

(b) Calculate the mid-span deflections of both beams assuming that they are subjected to a uniformly distributedload of 140 kN/m.

(A) LOAD CAPACITY

(i) Beam AElastic modulus of single 250 × 300 mm beam, Z s, is

Z

Iy

bdds

s3

3

/2mm= = =

×= ×

/

.

12 250 3006

3 75 102

6

Combined elastic modulus of two 250 × 300 mm beams acting separately, Z c = 2Z s = 7.5 × 106 mm3

Moment capacity of combined section, M, is

M = σZ c = 165 × 7.5 × 106 = 1.2375 × 109 N mm

Hence, load carrying capacity of Beam A, ωA, is

M

( )= =

×ω ωb2 3 2

88 10

8A = 1.2375 × 109 N mm

⇒ ωA = 154.7 N/mm = 154.7 kN/m

(ii) Beam BIn beam B the two smaller sections act together and behave like a beam 250 mm wide and 600 mm deep. The elasticmodulus of the combined section, Z, is

Z

Iy

bdd

/

= = =×

= ×3

3

/2mm

12 250 6006

15 102

6

Bending strength of beam, M = σZ = 165 × 15 × 106 = 2.475 × 106 N mmHence, load carrying capacity of Beam B, ωB, is

M

( ) . = =

×= ×

ω ωb2 3 29

88 10

82 475 10B N mm

⇒ ωB = 309.4 N/mm = 309.4 kN/m

9780415467193_C04 11/3/09, 11:10 AM200

201


smaller and hence cheaper foundations. One draw-back of this form of construction is that shutteringis needed and the slab propped until the concretedevelops adequate strength.

A development of this approach, which can elimi-nate the need for tensile steel reinforcement andpropping of the slab during construction, is to useprofiled metal decking as permanent shuttering(Fig. 4.35 ). Three common types of metal deckingused in composite slab construction are shownin Fig. 4.36. The metal decking is light and easyto work, which makes for simple and rapid con-struction. This system is most efficient for slabspans of between 3–4 m, and beam spans of up toaround 12 m. Where longer spans and/or higher

Fig. 4.35 Composite metal deck floor.

loads are to be supported, the steel beams may besubstituted with cellular beams or stub-girders (Fig.4.37 ). In structures where there is a need to reducethe depth of floor construction, for example tallbuildings, the Slimflor system developed by BritishSteel can be used. Floor spans are limited to 7.5 musing this system, which utilises stiff steel beams,fabricated from universal column sections weldedto a steel plate, and deep metal decking whichrests on the bottom flange of the beam (Fig. 4.38).

Nowadays, almost all composite floors in steelframed buildings are formed using profiled metaldecking and the purpose of the following sectionsis to discuss the design of (a) composite slabs and(b) composite beams.

4.10.1 COMPOSITE SLABSComposite slabs (i.e. metal decking plus concrete)are normally designed to BS 5950: Part 4. Althoughexplicit procedures are given in the standard, thesetend to be overly conservative when compared withthe results of full-scale tests. Therefore, designersmostly rely on load/span tables produced by metaldeck manufacturers in order to determine the thick-ness of slab and mesh reinforcement required for agiven floor arrangement, fire rating, method ofconstruction, etc. Table 4.17 shows an example ofa typical load/span table available from one sup-plier of metal decking. Concrete grades in the range30–40 N/mm2 are common. Slab depths may varybetween 100 and 200 mm. Example 4.20 illustratesthe use of this table.

4.10.2 COMPOSITE BEAMSOnce the composite slab has been designed, designof the primary and secondary composite beams (i.e.steel beams plus slab) can begin. This is normally

Example 4.16 continued(B) DEFLECTIONMid-span deflection of beam A (for a notional load of ω = 140 kN/m), δA, is

δ

ωA mm= =

× × ×× × × ×

= ( )

( . ) .

5384

5 140 8 10384 205 10 2 5 625 10

32 34 3 4

3 8

bEI

Mid-span deflection of beam B, δB, is

δ

ωB mm= =

× × ×× × × ×

= ( )

.

5384

5 140 8 10384 205 10 4 5 10

84 3 4

3 9

bEI

Hence, it can be seen that the load capacity has been doubled and the stiffness quadrupled by connecting the twobeams, a fact which will generally be found to hold for other composite sections.

9780415467193_C04 9/3/09, 1:40 PM201


202

11226

136

15 5570

Cover width 900 mm

(a)

40 112.5 152.5 137.5

Cover width 610 mm

Cover width 600 mmTrough shear-bond clip 56

210

87.5

21

38

(b)

(c)

51

15

Fig. 4.36 Steel decking: (a) re-entrant, (b) trapezoidal (c) deep deck.

Primarycompositebeam

Secondary composite beams

carried out in accordance with the recommenda-tions in Part 3: Section 3.1 of BS 5950, hereafter

referred to as BS 5950–3.1, and involves the follow-ing steps:

9780415467193_C04 9/3/09, 1:40 PM202

203

Concrete slab

Voids in web reduce self-weight andallow easy service distribution

(a)

Stud connectorsVoids Steel decking

Bottom chord Stub girdersSecondary beam

(b)


Fig. 4.37 Long span structures: (a) cellular beams (b) stub-girder.

Fig. 4.38 Slimflor system.

1. Determine the effective breadth of the concreteslab.

2. Calculate the moment capacity of the section.3. Evaluate the shear capacity of the section.4. Design the shear connectors.5. Assess the longitudinal shear capacity of the

section.6. Check deflection.

Each of these steps is explained below by refer-ence to simply supported secondary beams of class1 plastic UB section supporting a solid slab.

4.10.2.1 Effective breadth of concrete slab, BeAccording to BS 5950–3.1, the effective breadth ofconcrete slab, Be, acting compositely with simplysupported beams of length L should be taken asthe lesser of L/4 and the sum of the effectivebreadths, be, of the portions of flange each side ofthe centreline of the steel beam (Fig. 4.39).

9780415467193_C04 9/3/09, 1:40 PM203


204

Table 4.17 Typical load /span table for design of unpropped double span slab and deck made ofnormal weight grade 35 concrete (PMF, CF70, Corus).

Maximum span (m)

Deck thickness (mm)

Slab

0.9 1.2

FireDepth

Mesh

Total applied load (kN/m2)

Rating(mm)

3.50 5.00 10.0 3.50 5.00 10.0

1 hr 125 A142 3.2 3.2 2.8 4.0 3.7 3.0

11/2hr 135 A193 3.1 3.1 2.7 3.9 3.5 2.8

2 hr 150 A193 2.9 2.9 2.5 3.5 3.2 2.6200 A393 2.5 2.5 2.5 3.5 3.5 3.5250 A393 2.1 2.1 2.1 3.2 3.2 3.2

be be

Be is the lesser of:(a) beam span/4(b) 2be

Fig. 4.39 Effective breadth.

Fig. 4.40 Plastic neutral axis positions.

Case 1

Case 2

Case 3

d) The ultimate moment capacity of the com-posite section is independent of the method ofconstruction i.e. propped or unpropped.

The moment capacity of a composite section de-pends upon where the plastic neutral axis falls withinthe section. Three outcomes are possible, namely:

1. plastic neutral axis occurs within the concreteflange;

2. plastic neutral axis occurs within the steel flange;3. plastic neutral axis occurs within the web

(Fig. 4.40).

Only the first two cases will be discussed here.

(i) Case 1: Rc >>>>> Rs. Figure 4.41 shows the stressdistribution in a typical composite beam sectionwhen the plastic neutral axis lies within the con-crete slab.

Since there is no resultant axial force on thesection, the force in the concrete, R c′, must equalthe force in the steel beam, Rs. Hence

4.10.2.2 Moment capacityIn the analysis of a composite section to determineits moment capacity the following assumptions canbe made:

a) The stress block for concrete in compression atultimate conditions is rectangular with a designstress of 0.45fcu

b) The stress block for steel in both tension andcompression at ultimate conditions is rectangu-lar with a design stress equal to py

c) The tensile strength of the concrete is zero

9780415467193_C04 9/3/09, 1:41 PM204

205

Fig. 4.41 Stress distribution when plastic neutral axis lies within concrete flange.


Ds

D A Rs

py

Be

yp R ′c

0.45fcu

plastic neutralaxis

R c′ = R s (4.44)

whereR c′ = design stress in concrete × area of

concrete in compression= (0.45fcu)(Be yp) (4.45)

R s = steel design strength × area of steel section= py A

The maximum allowable force in the concreteflange, R c, is given by

Rc = design stress in concrete ×area of concrete flange

= (0.45fcu)(BeDs) (4.46)

Eliminating 0.45fcu from equations 4.45 and 4.46gives

R ′c =

R yDc p

s

(4.47)

Combining equations 4.44 and 4.47 and rearrang-ing obtains the following expression for depth ofthe plastic neutral axis, yp

y

RR

D Dps

cs s = ≤ (4.48)

Taking moments about the top of the concreteflange and substituting for yp, the moment capacityof the section, Mc, is given by

M R D

DR

yc s s c

p

2 2= +

− ′

= +

− ′ R DD

RRR

Ds s cs

cs22

= +

− R DD R D

Rs ss2

s

c2 2(4.49)

whereD = depth of steel sectionDs = depth of concrete flange

Note that the equation for Mc does not involveyp. Nonetheless it should be remembered that thisequation may only be used to calculate Mc pro-vided that yp < Ds. This condition can be checkedeither via equation 4.48 or, since Rc′ = Rs (equation4.44) and Rc > R c′, by checking that Rc > Rs. Clearlyif Rc < R s, it follows that the plastic neutral axisoccurs within the steel beam.

(ii) Case 2: Rc <<<<< Rs. Figure 4.42a shows the stressdistribution in the section when the plastic neutralaxis lies within the steel flange.

By equating horizontal forces, the depth ofplastic neutral axis below the top of the steel flange,y, is obtained as follows

Rc + py(By) = R s − py(By)

⇒ =

−

y

R RBp

s c

y2

Resistance of the steel flange, R f = py(BT ) ⇒ py

=

RBT

f

Substituting into the above expression for y gives

y

R RR T

T

=−

≤s c

f /

2(4.50)

The expression for moment capacity is derivedusing the equivalent stress distribution shown inFig. 4.42(b). Taking moments about the top of thesteel flange, the moment capacity of the section isgiven by

9780415467193_C04 9/3/09, 1:41 PM205


206

Ds

D Rs − pyBy

py

Be

Rc

T

B

Rs

y py

D/2

2py

Rc

(a) (b)

Fig. 4.42 Stress distribution when plastic neutral axis lies within steel flange.

Fig. 4.43 Slab thickness and depth of metal decking.

Ds

Section X−X

X

X

Dp

Omit fromconsideration

Profiledmetaldecking

M R

DR

Dp By

yc s c

sy2 2

( )= + − 22

= + − R

DR

Dp Bys c

sy2 2

2

Substituting for py and y and simplifying gives

M R

DR

D R RR

Tc s cs s c

f2 2= + −

( – )2

4(4.51)

(iii) Moment capacity of composite beamincorporating metal decking. The momentcapacity of composite beams incorporating profiledmetal decking is given by the following:

Case 1: Plastic neutral axis is in the concrete flange

M R

DD

RR

D Dc s s

s

c

s p

2 2

= + −

−

(4.52)

Case 2: Plastic neutral axis is in the steel flange

M R

DR

D D R RR

Tc s c

s p s c

f2 2 4

( )= +

+

−

− 2

(4.53)

These expressions are derived in the same wayas for beams incorporating solid slabs but furtherassume that (a) the ribs of the metal decking runperpendicular to the beams and (b) the concretewithin the depth of the ribs is ignored (Fig. 4.43).The stress distributions used to derive equations4.52 and 4.53 are shown in Fig. 4.44. Note thatthe symbols in these equations are as previouslydefined except for Rc, which is given by

Rc = 0.45fcuBe(Ds − Dp) (4.54)

where Dp is the overall depth of the profiled metaldecking (Fig. 4.43).

9780415467193_C04 9/3/09, 1:41 PM206

207

Ds

D

Rs

py

Be

yp R ′c

0.45fcu

Plasticneutralaxes

(a)

py

(b)

Rs

y

D/2

Ds − Dp

0.45fcu

Rc

2py

Fig. 4.44 Stress distributions in composite beams incorporating profiled metal decking: (a) plastic neutral axis is within theconcrete flange (b) plastic neutral axis is within the steel flange.


Shear studs are available in a range of diametersand lengths as indicated in Table 4.18. The 19 mmdiameter by 100 mm high stud is by far the mostcommon in buildings. In slabs comprising profiledmetal decking and concrete, the heights of the studsshould be at least 35 mm greater than the overalldepth of the decking. Also, the centre-to-centredistance between studs along the beam should liebetween 5φ and 600 mm or 4Ds if smaller, whereφ is the shank diameter and Ds the depth of theconcrete slab. Some of the other code recommen-dations governing the minimum size, transversespacing and edge distances of shear connectors areshown in Fig. 4.45.

(ii) Design procedure. The shear strength ofheaded studs can be determined using standardpush-out specimens consisting of a short sectionof beam and slab connected by two or four studs.Table 4.18 gives the characteristic resistances, Qk,of headed studs embedded in a solid slab ofnormal weight concrete. For positive (i.e. sagging)moments, the design strength, Qp, should betaken as

Qp = 0.8Qk (4.56)

A limit of 80% of the static capacity of the shearconnector is deemed necessary for design to ensurethat full composite action is achieved between theslab and the beam.

The capacity of headed studs in composite slabswith the ribs running perpendicular to the beamshould be taken as their capacity in a solid slab

4.10.2.3 Shear capacityAccording to BS 5950–3.1, the steel beam shouldbe capable of resisting the whole of the verticalshear force, Fv. As discussed in 4.8.5, the shearcapacity, Pv, of a rolled I-section is given by

Pv = 0.6pytD

Like BS 5950: Part 1, BS 5950–3.1 recommendsthat where the co-existent shear force exceeds 0.5Pv

the moment capacity of the section, Mcv, should bereduced in accordance with the following:

Mcv = Mc − (Mc − M f)(2Fv/Pv − 1)2 (4.55)

whereMf is the plastic moment capacity of that part

of the section remaining after deduction ofthe shear area Av defined in Part 1 ofBS 5950

Pv is the lesser of the shear capacity and theshear buckling resistance, both determinedfrom Part 1 of BS 5950

4.10.2.4 Shear connectors

(i) Headed studs. For the steel beams and slabto act compositely and also to prevent separationof the two elements under load, they must be struc-turally tied. This is normally achieved by providingshear connectors in the form of headed studs asshown in Fig. 4.45. The shear studs are usuallywelded to the steel beams through the metaldecking.

9780415467193_C04 9/3/09, 1:42 PM207


208

Fig. 4.45 Geometrical requirements for placing of studs (CIRIA Report 99).

3ø h

50 mm

ø

50 mm

20 mm

Dp

D s

35 mm

15 mm

0.4ø

1.5ø

4ø25 mm or 1.25ø

Table 4.18 Characteristic resistance, Qk, of headed studs in normalweight concrete (Table 5, BS 5950–3.1)

Shank diameter Height Characteristic strength (N/mm2)(mm) (mm)

25 30 35 40

25 100 146 154 161 168 kN22 100 119 126 132 139 kN19 100 95 100 104 109 kN19 75 82 87 91 96 kN16 75 70 74 78 82 kN13 65 44 47 49 52 kN

For full shear connection, the total number ofstuds, Np, required over half the span of a simplysupported beam in order to develop the positivemoment capacity of the section can be determinedusing the following expression:

Np = Fc /Qp (4.57)

whereFc = Apy (if plastic neutral axis lies in the

concrete flange)Fc = 0.45fcuBeDs (if plastic neutral axis lies in the

steel beam)Qp = design strength of shear studs = 0.8Q k

(in solid slab) and kQ k (in a composite slabformed using ribbed profile sheeting alignedperpendicular to the beam)

multiplied by a reduction factor, k, given by thefollowing expressions:

for one stud per ribk = 0.85(br /Dp){(h/Dp) − 1} ≤ 1

for two studs per ribk = 0.6(br /Dp){(h/Dp) − 1} ≤ 0.8

for three or more studs per ribk = 0.5(br /Dp){(h/Dp) − 1} ≤ 0.6

wherebr breadth of the concrete ribDp overall depth of the profiled steel sheeth overall height of the stud but not more than

2Dp or Dp + 75 mm, although studs ofgreater height may be used

9780415467193_C04 9/3/09, 1:43 PM208

209


In composite floors made with metal decking itmay not always be possible to provide full shear con-nection between the beam and the slab becausethe top flange of the beam may be too narrowand/or the spacing of the ribs may be too great to

accommodate the total number of studs required. Inthis case, the reader should refer to BS 5950–3.1:Appendix B, which gives alternative expressions formoment capacity of composite sections with partialshear connection.

(A) PLASTIC NEUTRAL AXIS OF COMPOSITE SECTIONResistance of the concrete flange, Rc , is

Rc = (0.45fcu)BeDs = (0.45 × 35)1500 × 130 × 10−3 = 3071.25 kN

From Table 4.3, since T (= 16 mm) ≤ 16 mm and steel grade is S275, design strength of beam, py = 275 N/mm2.Resistance of steel beam, R s, is

R s = Apy = 95 × 102 × 275 × 10−3 = 2612.5 kN

Since Rc > R s, the plastic neutral axis falls within the concrete slab. Confirm this by calculating yp:

y

Apf Bp

y

cu e(0.45

)

. .

= =×

× ×2612 5 10

0 45 35 1500

3

= 110.6 mm < Ds OK

(B) MOMENT CAPACITYSince yp < Ds use equation 4.49 to calculate the moment capacity of the section, Mc. Hence

M Ap D

D RR

Dc y s

s

c

s

2 2kN m .

.

..

. = + −

= × + − ×

=−2612 5 10 130412 8

22612 53071 25

1302

10 734 43 6

Fig. 4.46

Example 4.17 Moment capacity of a composite beam (BS 5950)Determine the moment capacity of the section shown in Fig. 4.46 assuming the UB is of grade S275 steel and thecharacteristic strength of the concrete is 35 N/mm2.

1500 mm

130 mm

406 × 178 × 74 UB

9780415467193_C04 9/3/09, 1:43 PM209


210

Example 4.18 Moment capacity of a composite beam (BS 5950)Repeat Example 4.17 assuming the beam is made of grade S355 steel. Also, design the shear connectors assuming thebeam is 6 m long and that full composite action is to be provided.

PLASTIC NEUTRAL AXIS OF COMPOSITE SECTIONAs before, the resistance of the concrete flange, Rc, is

R c = (0.45fcu)A c = (0.45 × 35)1500 × 130 × 10−3 = 3071.25 kN

From Table 4.3, since T (= 16 mm) ≤ 16 mm and steel grade is S355, design strength of beam, py = 355 N/mm2.Resistance of steel beam, R s, is

R s = Apy = 95 × 102 × 355 × 10−3 = 3372.5 kN

Since Rc < R s, the plastic neutral axis will lie within the steel beam. Confirm this by calculating y :

y

R RBp

. .

. . =

−=

× − ×× ×

=s c

y2mm

3372 5 10 3071 25 102 179 7 355

2 43 3

OK

MOMENT CAPACITYSince y < T use equation 4.51 to calculate moment capacity of the section, Mc.

Resistance of steel flange, R f = BTpy = 179.7 × 16 × 355 = 1.02 × 106 N

Moment capacity of composite section, Mc, is

M R

DR

D R RR

Tc s c

s s c

f2 2

( )= + −

− 2

4

= × + × −

− ××

. .

. (( . . ) )

. 3372 5 10

412 82

3071 25 10130

23372 5 3071 25 10

1 02 10164

3 33 2

6

= 895.4 × 106 N mm = 895.4 kN m

SHEAR STUDSFrom Table 4.18, characteristic resistance, Q k, of headed studs 19 mm diameter × 100 mm high embedded in grade35 concrete is 104 kN.Design strength of studs under positive moment, Qp, is given by

Qp = 0.8Q k = 0.8 × 104 = 83.2 kN

Number of studs required = = ≥

..

RQ

c

p

36.93071 25

83 2Provide 38 studs, evenly arranged in pairs, in each half span of beam as shown.

225

X

X

3 m

19 pairs of studs @ 150 centres

Centre line of beam

Section X–X

75

9780415467193_C04 9/3/09, 1:43 PM210

211

4.10.2.5 Longitudinal shear capacityA solid concrete slab, which is continuous over sup-ports, will need to be reinforced with top and bottomsteel to resist the sagging and hogging moments dueto the applied loading. Over beam supports, this steelwill also be effective in transferring longitudinal forcesfrom the shear connectors to the slab without splitt-ing the concrete. Design involves checking that theapplied longitudinal shear force per unit length, ν, doesnot exceed the shear resistance of the concrete, νr.

The total longitudinal shear force per unit length,ν, is obtained using the following

ν = NQp/s (4.58)

whereN number of shear connectors in a groups longitudinal spacing centre-to-centre of

groups of shear connectorsQp design strength of shear connectors

In a solid slab, the concrete shear resistance, νr, isobtained using the following:

νr = 0.7Asv fy + 0.03ηAcv fcu ≤ 0.8ηAcv fcu (4.59)

In slabs with profiled steel sheeting, νr, is given by

νr = 0.7Asv fy + 0.03ηAcv fcu + νp

≤ 0.8ηAcv fcu + νp (4.60)

wherefcu characteristic strength of the concrete

≤ 40 N/mm2

η 1.0 for normal weight concreteAcv mean cross-sectional area per unit length of

the beam of the concrete surface underconsideration

Asv cross-sectional area, per unit length of thebeam, of the combined top and bottomreinforcement crossing the shear surface(Fig. 4.47)

νp contribution of the profiled steel decking.Assuming the decking is continuous acrossthe top flange of the steel beam and that theribs are perpendicular to the span of thebeam, νp is given by

νp = tppyp (4.61)

in whichtp thickness of the steel deckingpyp design strength of the steel decking obtained

either from Part 4 of BS 5950 ormanufacturer’s literature

4.10.2.6 DeflectionThe deflection experienced by composite beamswill vary depending on the method of construction.Thus where steel beams are unpropped duringconstruction, the total deflection, δT, will be thesum of the dead load deflection, δD, due to the selfweight of the slab and beam, based on the proper-ties of the steel beam alone, plus the imposed loaddeflection, δ I, based on the properties of the com-posite section:

δT = δD + δ I

For propped construction the total deflection iscalculated assuming that the composite section sup-ports both dead and imposed loads.

The mid-span deflection of a simply supportedbeam of length L subjected to a uniformly distrib-uted load, ω, is given by

Fig. 4.47

(a) Solid slab

a

ab b

Ab

A t

e

e

A t

Sheeting

(b) Composite slab with the sheetingspanning perpendicular to the beam

Surface Asv

a–ab–be–e

Ab + A t

2Ab

A t


9780415467193_C04 9/3/09, 1:44 PM211


212

δ

ω =

5 LEI

4

384

whereE elastic modulus = 205 kN/mm2 for steelI second moment of area of the section

Deflections of simply supported composite beamsshould be calculated using the gross value of thesecond moment of area of the uncracked section,Ig, determined using a modular ratio approach. Theactual value of modular ratio, αe, depends on theproportions of the loading which are considered tobe long term and short term. Imposed loads onfloors should be assumed to be 2/3 short term and1/3 long term. On this basis, appropriate values ofmodular ratio for normal weight and lightweightconcrete are 10 and 15, respectively. For com-posite beams with solid slabs Ig is given by

I I

B D AB D D DA B Dg s

e s3

e

e s s

e e s

= + ++

+

( )( )12 4

2

α α(4.62)

where Is is the second moment of area of the steelsection.

Equation 4.62 is derived assuming that the con-crete flange is uncracked and unreinforced (seeAppendix D). In slabs with profiled steel sheetingthe concrete within the depth of the ribs may con-servatively be omitted and Ig is then given by:

I I

B D Dg s

e s p

e

( )

= +− 3

12α

+

− + ++ −

( )( )

( [ ])

AB D D D D D

A B D De s p s p

e e s p

2

4 α(4.63)

The deflections under unfactored imposed loadsshould not exceed the limits recommended inBS 5590, as summarised in Table 4.5.

Example 4.19 Design of a composite floor (BS 5950)Steel UBs at 3.5 m centres with 9 m simple span are to support a 150 mm deep concrete slab of characteristicstrength 30 N/mm2 (Fig. 4.48). If the imposed load is 4 kN/m2 and the weight of the partitions is 1 kN/m2

a) select a suitable UB section in grade S355 steelb) check the shear capacityc) determine the number and arrangement of 19 mm diameter × 100 mm long headed stud connectors requiredd) assuming the slab is reinforced in both faces with H8@150 centres (A = 335 mm2/m), check the longitudinal

shear capacity of the concretee) calculate the imposed load deflection of the beam.

Assume that weight of the finishes, and ceiling and service loads are 1.2 kN/m2 and 1 kN/m2 respectively. The densityof normal weight reinforced concrete, ρc, can be taken as 24 kN/m3.

BEAM SELECTION

Design momentBeam span, L = 9 mSlab span, b = 3.5 mDesign load per beam, ω = 1.4(Dsρc + finishes + ceiling/services)b + 1.6(qk + partition loading)b

= 1.4(0.15 × 24 + 1.2 + 1)3.5 + 1.6(4 + 1)3.5 = 56.4 kN/m

Design moment, M

L

. = =

×=

ω 2

8kN m

56 4 98

5712

Fig. 4.48

3.5 m 3.5 m 3.5 m

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213


Effective width of concrete slabBe is the lesser of beam span/4 (= 9000/4 = 2250 mm) and beam spacing (= 3500 mm)∴ Be = 2250 mm

Moment capacityUsing trial and error, try 356 × 171 × 45 UB in grade S355 steelResistance of the concrete flange, Rc, is

Rc = (0.45fcu)BeDs = (0.45 × 30) 2250 × 150 × 10−3 = 4556.3 kN

From Table 4.3, since T (= 9.7 mm) < 16 mm and steel grade is S355, design strength of beam, py = 355 N/mm2.Resistance of steel beam, R s, is

R s = Apy = 57 × 102 × 355 × 10−3 = 2023.5 kN

Since Rc > R s, the plastic neutral axis will lie in the concrete slab. Confirm this by substituting into the followingexpression for yp

⇒ = =

×× ×

= (0.45

mm < py

cu esy

Apf B

D )

.

. .

2023 5 100 45 30 2250

66 63

OK

Hence, moment capacity of composite section, Mc, is

M Ap D

D RR

Dc y s

s

c

s

2kN m .

.

. . = + −

= × + − ×

=−

22023 5 10 150

3522

2023 54556 3

1502

10 592 33 6

Mc > M + M sw = 571 + (45 × 9.8 × 10−3)92/8 = 575.5 kN m OK

SHEAR CAPACITY

Shear force, Fv =

12

12

ωL = × 56.4 × 9 = 253.8 kN

Shear resistance, Pv = 0.6pyt D = 0.6 × 355 × 6.9 × 352 × 10−3 = 517 kN > Fv OKAt mid-span, Fv (= 0) < 0.5Pv (= 258 kN) and therefore the moment capacity of the section calculated above is valid.

SHEAR CONNECTORSFrom Table 4.18, characteristic resistance, Q k, of headed studs 19 mm diameter × 100 mm high is 100 kN.Design strength of shear connectors, Qp, is

Qp = 0.8Q k = 0.8 × 100 = 80 kN

Longitudinal force that needs to be transferred, Fc, is 2023.5 kN

Number of studs required = = ≥

.

RQ

c

p

25.32023 5

80

Provide 26 studs, evenly arranged in pairs, in each half span of beam.

Spacing =

450012

= 375 mm, say 350 mm centres


12512 × 350 = 4200 (26 studs)

175

4500 mmCentre lineof beam

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214

Shear failure surface a–aLength of failure surface = 2 × 150 = 300 mmCross-sectional area of failure surface per unit length of beam, A cv = 300 × 103 mm2/mCross-sectional area of reinforcement crossing potential failure surface, A sv, is

A sv = A t + Ab = 2 × 335 = 670 mm2/m

Hence shear resistance of concrete, υr, is

υr = 0.7A sv fy + 0.03ηA cv fcu ≤ 0.8ηA cv fcu = ( . . ( )0 8 1 0 300 10 303× × × /103 = 1315 N/mm

= ( . . . )0 7 670 500 0 03 1 0 300 10 303× × + × × × × /103 = 504.5 N/mm > υ OK

Shear failure surface b–bLength of failure surface = 2 times stud height + 7φ = 2 × 100 + 7 × 19 = 333 mmCross-sectional area of failure surface per unit length of beam, A cv = 333 × 103 mm2/mCross-sectional area of reinforcement crossing potential failure surface, A sv, is

A sv = 2Ab = 2 × 335 = 670 mm2/mm


υr = 0.7A sv fy + 0.03ηA cv fcu ≤ 0.8ηAcv fcu = ( . . ( ) )0 8 1 0 333 10 303× × × /103 1459 N/m

= ( . . . )0 7 670 500 0 03 1 0 333 10 303× × + × × × × /103 = 534 N/mm > υ OK

DEFLECTIONSince beam is simply supported use the gross value of second moment of area, Ig, of the uncracked section tocalculate deflections.

I I


e s3

e

e s s2

e e s12( )

= + ++

+

( )α α4

= × +

××

+× × × +

× × + ×= ×

( )( )

12100 102250 150

12 1057 10 2250 150 352 150

4 57 10 10 2250 15049150 104

3 2 2

24 mm4

LONGITUDINAL SHEARLongitudinal force, υ, is

υ

= =

× ×=

NQs

p N/mm2 80 10

350457

3

Ds = 150

a

a

b b

100

1.5c 4c 1.5ca

a

At = 335 mm2/m

Ab = 335 mm2/m

c = shank diameterof stud


9780415467193_C04 9/3/09, 1:46 PM214

215


Example 4.20 Design of a composite floor incorporating profiled metaldecking (BS 5950)

Figure 4.49 shows a part plan of a composite floor measuring 9 m × 6 m. The slab is to be constructed using profiledmetal decking and normal weight, grade 35 concrete and is required to have a fire resistance of 60 mins. The lon-gitudinal beams are of grade S355 steel with a span of 9 m and spaced 3 m apart. Design the composite slab andinternal beam A2–B2 assuming the floor loading is as follows:

imposed load = 4 kN/m2

partition load = 1 kN/m2

weight of finishes = 1.2 kN/m2

weight of ceiling and services = 1 kN/m2

The density of normal weight reinforced concrete can be taken to be 24 kN/m3.

Example 4.19 continuedMid-span deflection of beam, δ, is

δ

ω

( . )

= =× × × ×

× × × ×5 5 5 3 5 9 9 10

384 205 10 49 150 10

3 12

3 4

LEI

4

384

= = = . 14 8

9000360

25mm < 360

mmL

OK

Therefore adopt 356 × 171 × 45 UB in grade S355 steel.

SLAB DESIGNAssuming the slab is unpropped during construction, use Table 4.17 to select suitable slab depth and deck gauge forrequired fire resistance of 1 hr. It can be seen that for a total imposed load of 7.2 kN/m2 (i.e. occupancy, partitionload, finishes, ceilings and services) and a slab span of 3 m, a 125 mm thick concrete slab reinforced with A142 meshand formed on 1.2 mm gauge decking should be satisfactory.

3

2

1A B

9 m

3 m

3 m

Fig. 4.49 Part plan of composite floor.

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216

Example 4.20 continuedA cross-section through the floor slab is shown below.

Dp = 55

Ribs at 300 c/c112

26

149 A142 Mesh

Ds = 125

BEAM A2–B2Beam selectionDesign moment

Beam span, L = 9 mBeam spacing, b = 3 m

From manufacturers’ literature effective slab thickness, Def = Ds − 26 = 125 − 26 = 91 mm

Design load, ω = 1.4(Defρc + finishes + ceiling/services)b + 1.6(q k + partition loading)b

= 1.4(0.091 × 24 + 1.2 + 1)3 + 1.6(4 + 1)3 = 42.4 kN/m

Design moment, M

L

8kN m

2

= =×

= .

. ω 42 4 9

8429 3

2

Effective width of concrete slabBe is the lesser of L/4 (= 9000/4 = 2250 mm) and beam spacing (= 3000 mm)∴ Be = 2250 mm

Moment capacityUsing trial and error, try 305 × 165 × 40 UB in grade S355 steelResistance of concrete flange, R c, is

R f B D Dc cu e s p kN ( . ) ( ) ( . ) ( ) . = − = × × × × × =−0 45 0 45 35 2250 125 55 10 2480 63

From Table 4.3, since T (= 10.2 mm) < 16 mm and steel grade is S355, design strength of beam, p y = 355 N/mm2.Resistance of steel beam, R s, is

R s = Apy = 51.5 × 102 × 355 × 10−3 = 1828.3 kN

Since R c > R s, plastic neutral axis lies within the slab. Confirm this by substituting into the following expression for yp

y

Apf Bp

y

cu e0.45= =

×× ×

( )

.

. 1828 3 10

0 45 35 2250

3

= 51.6 mm < Ds − Dp = 125 − 55 = 70 mm OK

Hence, moment capacity of composite section, Mc, is

M R

DD

RR

D Dc s s

s

c

s p

2 2

= + −

−

= × + −

××

−

× =− .

.

.

.

1828 3 10303 8

2125

1828 3 102480 6 10

125 552

10 45933

36 kN m

Mc > M + Msw = 429.3 + 1.4(40 × 9.8 × 10−3)92/8 = 434.9 kN m OK

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217


Shear capacity

Shear force, Fv =

12

12

ωL = × 42.4 × 9 = 190.8 kN

Shear resistance, Pv = 0.6pytD = 0.6 × 355 × 6.1 × 303.8 × 10−3 = 394.7 kN > Fv OKAt mid-span Fv = 0 < 0.5Pv (= 197.4 kN). Therefore moment capacity of section remains unchanged.

Shear connectorsAssume headed studs 19 mm diameter × 100 mm high are to be used as shear connectors.From Table 4.18, characteristic resistance of studs embedded in a solid slab, Q k = 104 kNDesign strength of studs under positive (i.e. sagging) moment, Qp, is

Qp = 0.8Q k = 0.8 × 104 = 83.2 kN

Design strength of studs embedded in a slab comprising profiled metal decking and concrete, Q p′ , is

Q p′ = kQ p

Assuming two studs are to be provided per trough, the shear strength reduction factor, k, is given by

k = 0.6(br/Dp) {(h/Dp) − 1} ≤ 0.8

= 0.6(149/55){(95/55) − 1} = 1.2

∴ k = 0.8

⇒ Q p′ = 0.8 × 83.2 = 66.6 kN

Maximum longitudinal force in the concrete, Fc = 1828.3 kN

Total number of studs required =

RQ

c

p

.

. .= =

1828 366 6

27 5

As the spacing of deck troughs is 300 mm, fifteen (= 4500/300) trough positions are available for fixing of the shearstuds. Therefore, provide 30 studs in each half span of beam, i.e. two studs per trough.

Longitudinal shearLongitudinal shear stressLongitudinal force, υ, is

υ

. = =

× ×=

NQs

p N/mm2 66 6 10

300444

3

Shear failure surface e–eCross-sectional area of reinforcement crossing potential failure surface, A sv, is

A sv = A t = 142 mm2/m

Ds = 125

e

e

At = 142 mm2/m

55

Acv

112

164

300


9780415467193_C04 9/3/09, 1:47 PM217


218

connections in steel structures. However, at theoutset, it is worthwhile reiterating some generalpoints relating to connection design given in clause6 of BS 5950.

The first couple of sentences are vitally import-ant – ‘Joints should be designed on the basis ofrealistic assumptions of the distribution of internalforces. These assumptions should correspond withdirect load paths through the joint, taking accountof the relative stiffnesses of the various componentsof the joint’. Before any detailed design is embarkedupon therefore, a consideration of how forces willbe transmitted through the joint is essential.

‘The connections between members should becapable of withstanding the forces and moments towhich they are subject . . . without invalidating thedesign assumptions’. If, for instance, the structureis designed in ‘simple construction’, the beam-column joint should be designed accordingly toaccept rotations rather than moments. A rigid jointwould be completely wrong in this situation, as it

4.11 Design of connectionsThere are two principal methods for connectingtogether steel elements of structure, and the variouscleats, end plates, etc. also required.

1. Bolting, using ordinary or high strength frictiongrip (HSFG) bolts, is the principal method ofconnecting together elements on site.

2. Welding, principally electric arc welding, is analternative way of connecting elements on site,but most welding usually takes place in factoryconditions. End plates and fixing cleats arewelded to the elements in the fabrication yard.The elements are then delivered to site wherethey are bolted together in position.

Figure 4.50 shows some typical connections usedin steel structures.

The aim of this section is to describe the designof some commonly used types of bolted and welded

Mean cross-sectional area of shear surface, A cv, is

[125 × 300 − 1/2(164 + 112)55]/0.3 = 99.7 × 103 mm2/m


υr = 0.7A sv fy + 0.03η A cv fcu + υp ≤ 0.8ηA cv fcu + υp

= × × + × × × × + × × ( . . . . . )0 7 142 500 0 03 1 0 99 7 10 35 1 2 10 2803 3 /103 = 490 N/mm

≤ ( . . . . )0 8 1 0 99 7 10 35 1 2 10 2803 3× × × + × × /103 = 808 N/mm OK

υr = 490 N/mm > υ OK

(Note pyp = 280 N/mm2 from the steel decking manufacturer’s literature.)

DeflectionSince beam is simply supported use the gross value of second moment of area, Ig, of the uncracked section tocalculate deflection.

I I

B D D AB D D D D DA B D Dg s

e s p

e

e s p s p

e e s p

( )

( )( ){ ( )}

= +−

+− + +

+ −

3 2

12 4α α

= × +

−×

+× × − + +

× × + −

( )

. ( )( . )

{ . ( )}8520 10

2250 125 5512 10

51 5 10 2250 125 55 303 8 125 554 51 5 10 10 2250 125 55

43 2 2

2

= 31873 × 104 mm4

Mid-span deflection of beam, δ, is

δ

ω

( )

. = =× × × ×

× × × ×= = =

5 5 5 3 9 9 10384 205 10 31 873 10

19 69000360

253 12

3 4

LEI

L4

384mm <

360mm OK

Therefore adopt 305 × 165 × 40 UB in grade S355 steel.


9780415467193_C04 9/3/09, 1:47 PM218

219

Design of connections

Fig. 4.50 Typical connections: (a) beam to column; (b) beam to beam.

Boltedweb cleat

Boltedend-plate

(a)

Boltedend-plate

Flangecover plate Web plate

(b)

Fig. 4.51 Rules for fastener spacing and edge/end distancesto fasteners.

p e2

e1

Direction

of stress

Dh

Holediameter

would tend to generate a moment in the columnfor which it has not been designed.

‘The ductility of steel assists in the distributionof forces generated within a joint.’ This meansthat residual forces due to initial lack of fit, or dueto bolt tightening, do not normally have to beconsidered.

4.11.1 BOLTED CONNECTIONSAs mentioned above, two types of bolts commonlyused in steel structures are ordinary (or black) boltsand HSFG bolts. Black bolts sustain a shear loadby the shear strength of the bolt shank itself,whereas HSFG bolts rely on a high tensile strengthto grip the joined parts together so tightly that theycannot slide.

There are three grades of ordinary bolts, namely4.6, 8.8 and 10.9. HSFG bolts commonly used instructural connections conform to the general gradeand may be parallel shank fasteners designed to benon-slip in service or waisted shank fasteners de-signed to be non-slip under factored loads. Thepreferred size of steel bolts are 12, 16, 20, 22, 24and 30 mm in diameter. Generally, in structuralconnections, grade 8.8 bolts having a diameter notless than 12 mm are recommended. In any case, asfar as possible, only one size and grade of boltshould be used on a project.

The nominal diameter of holes for ordinary bolts,Dh, is equal to the bolt diameter, d b, plus 1 mm for12 mm diameter bolts, 2 mm for bolts between 16and 24 mm in diameter and 3 mm for bolts 27mm or greater in diameter (Table 33: BS 5950):

Dh = d b + 1 mm for d b = 12 mm

Dh = d b + 2 mm for 16 ≤ d b ≤ 24 mm

Dh = d b + 3 mm for d b ≥ 27 mm

4.11.2 FASTENER SPACING ANDEDGE/END DISTANCES

Clause 6.2 of BS 5950 contains various recom-mendations regarding the distance between fas-teners and edge/end distances to fasteners, some ofwhich are illustrated in Fig. 4.51 and summarisedbelow:

1. Spacing between centres of bolts, i.e. pitch (p),in the direction of stress and not exposed tocorrosive influences should lie within the follow-ing limits:

2.5d b ≤ p ≤ 14t

where d b is the diameter of bolts and t the thick-ness of the thinner ply.

2. Minimum edge distance, e1, and end distance,e2, to fasteners should conform with the follow-ing limits:

Rolled, machine flame cut, sawn or planededge/end ≥ 1.25Dh

Sheared or hand flame cut edge/end ≥ 1.40Dh

9780415467193_C04 9/3/09, 1:47 PM219


220

where Dh is the diameter of the bolt hole. Notethat the edge distance, e1, is the distance fromthe centre line of the hole to the outside edge ofthe plate at right angles to the direction of thestress, whereas the end distance, e2, is the dis-tance from the centre line to the edge of theplate in the direction of stress.

3. Maximum edge distance, e1, should not exceedthe following:

e1 ≤ 11tε

where t is the thickness of the thinner part andε = (275/py)

1/2.

4.11.3 STRENGTH CHECKSBolted connections may fail due to various mechan-isms including shear, bearing, tension and combinedshear and tension. The following sections describethese failure modes and outline the associateddesign procedures for connections involving (a)ordinary bolts and (b) HSFG bolts.

4.11.3.1 Ordinary bolts

Shear and bearing. Referring to the connectiondetail shown in Fig. 4.52, it can be seen that theloading on bolt A between the web cleat and thecolumn will be in shear, and that there are threeprincipal ways in which the joint may fail. Firstly,the bolts can fail in shear, for example along sur-face x1–y1 (Fig. 4.52(a)). Secondly the bolts canfail in bearing as the web cleat cuts into the bolts(Fig. 4.52(b)). This can only happen when thebolts are softer than the metal being joined. Thirdly,the metal being joined, i.e. the cleat, can fail inbearing as the bolts cut into it (Fig. 4.52(c)). Thisis the converse of the above situation and can onlyhappen when the bolts are harder than the metalbeing joined.

It follows, therefore, that the design shear strengthof the connection should be taken as the least of:

1. Shear capacity of the bolt,

Ps = psAs (4.64)

2. Bearing capacity of bolt,

Pbb = d btp pbb (4.65)

3. Bearing capacity of connected part,

Pbs = kbsdbtppbs ≤ 0.5kbsetppbs (4.66)

whereps shear strength of the bolts (Table 4.19)pbb bearing strength of the bolts (Table 4.20)

P

Beamshear

A

P

(a)

y1

x1

(b)

(c)

Section P–P

Fig. 4.52 Failure modes of a beam-to-column connection:(a) single shear failure of bolt; (b) bearing failure of bolt;(c) bearing failure of cleat.

p bs bearing strength of the connected part(Table 4.21)

e end distance e2

A s effective area of bolts in shear, normally takenas the tensile stress area, A t (Table 4.22)

tp thickness of connected partkbs = 1.0 for bolts in standard clearance holes

Double shear. If a column supports two beams inthe manner indicated in Fig. 4.53, the failure modes

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221

Table 4.19 Shear strength of bolts (Table 30,BS 5950)

Bolt grade Shear strengthps (N/mm2)

4.6 1608.8 375

10.9 400General grade HSFG ≤ M24 400to BS 4395–1 ≥ M27 350Higher grade HSFG to BS 4395–2 400Other grades (Ub ≤ 1000 N/mm2) 0.4Ub

Note. Ub is the specified minimum tensile strength of the bolt.


Table 4.20 Bearing strength of bolts (Table 33,BS 5950)

Bolt grade Bearing strengthpbb (N/mm2)

4.6 4608.8 1000

10.9 1300General grade HSFG ≤ M24 1000to BS 4395–1 ≥ M27 900Higher grade HSFG to BS 4395–2 1300Other grades (Ub ≤ 1000 N/mm2) 0.7(Ub + Yb)

Note. Ub is the specified minimum tensile strength of the boltand Yb is the specified minimum yield strength of the bolt.

Table 4.21 Bearing strength of connected parts (Table 32)

Steel grade S275 S355 S460 Other gradesBearing strength pbs (N/mm2) 460 550 670 0.67(Ub + Yb)

Note. Ub is the specified minimum tensile strength of the bolt and Yb is the specifiedminimum yield strength of the bolt.

Table 4.22 Tensile stress area, At

Nominal size and thread Tensile stress area, At

diameter (mm) (mm2)

12 84.316 15720 24522 30324 35327 45930 561

Tension. Tension failure may arise in simple con-nections as a result of excessive tension in the bolts(Fig. 4.54(a)) or cover plates (Fig. 4.54(b)). Thetension capacity of ordinary bolts may be calcu-lated using a simple or more exact approach. Onlythe simple method is discussed here as it is botheasy to use and conservative. The reader is re-ferred to clause 6.3.4.3 of BS 5950 for guidanceon the more exact method.

According to the simple method, the nominaltension capacity of the bolt, Pnom, is given by

Pnom = 0.8pt A t (4.68)

wherept tension strength of the bolt (Table 4.23)A t tensile stress area of bolt (Table 4.22)

The tensile capacity of a flat plate is given by

Pt = αe py (4.69)

where effective net area, αe, is

αe = Keαn < αg (4.70)

in whichKe = 1.2 for grade S275 steel platesαg gross area of plate = bt (Fig. 4.54)αn net area of plate = αg – allowance for bolt

holes (= Dht, Fig. 4.54(b)).

essentially remain the same as for the previous case,except that the bolts (B) will be in ‘double shear’.This means that failure of the bolts will only occuronce surfaces x2–y2 and x3–y3 exceed the shearstrength of the bolt (Fig. 4.53(b)).

The shear capacity of bolts in double shear, Psd,is given by

Psd = 2Ps (4.67)

Thus, double shear effectively doubles the shearstrength of the bolt.

9780415467193_C04 9/3/09, 1:49 PM221


222

Fig. 4.53 Double shear failure.

Beamshear

(a)

y2

(b)

Section S–S

Beamshear

S

S

B

Top cleats

y3

x2 x3

Table 4.23 Tensile strength of bolts (Table 34,BS 5950)

Bolt grade Tension strengthpt (N/mm2)

4.6 2408.8 560

10.9 700General grade HSFG ≤ M24 590to BS 4395–1 ≥ M27 515Higher grade HSFG to BS 4395–2 700Other grades (Ub ≤ 1000 N/mm 2) 0.7Ub but ≤ Yb

Note. Ub is the specified minimum tensile strength of the boltand Yb is the specified minimum yield strength of the bolt.

Combined shear and tension. Where ordinarybolts are subject to combined shear and tension(Fig. 4.55 ), in addition to checking their shear andtension capacities separately, the following relation-ship should also be satisfied:

FP

FP

s

s

t

nom

.+ ≤ 1 4 (4.71)

whereFs applied shearFt applied tensionPs shear capacity (equation 4.64)Pnom tension capacity (equation 4.68).

Note that this expression should only be used whenthe bolt tensile capacity has been calculated usingthe simple method.

Fig. 4.54 Typical tension failures: (a) bolts in tension; (b) cover plate in tension.

(a)

Dh

(b)

b

Rupture

tt

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223

The slip factor depends on the condition of thesurfaces being joined. According to Table 35 ofBS 5950, shot or grit blasted surfaces have a slipfactor of 0.5 whereas wire brushed and untreated sur-faces have slip factors of 0.3 and 0.2 respectively.

Bearing. The bearing capacity of connected partsafter slip, Pbg, is given by

Pbg = 1.5dbtppbs ≤ 0.5etp pbs (4.74)

whered b bolt diametertp thickness of connected partpbs bearing strength of connected parts (Table 4.21)e end distance

Shear. As in the case of black bolts, the shearcapacity of HSFG bolts, Ps, is given by

Ps = psAs (4.75)

whereps shear strength of the bolts (Table 4.19)As effective area of bolts in shear, normally

taken as the tensile stress area, A t

(Table 4.22)

Combined shear and tension. When parallelshank friction grip bolts designed to be non-slip inservice, are subject to combined shear and tension,then the following additional check should becarried out:

FP

FP

s

sL

tot

o

.

+ ≤1 1

1 but Ftot ≤ At pt (4.76)

whereFs applied shearFtot total applied tension in the bolt, including

the calculated prying forces = ptA t in whichpt is obtained from Table 4.23 and A t isobtained from Table 4.22

PsL slip resistance (equation 4.72)Po minimum shank tension (Table 4.24 ).

4.11.4 BLOCK SHEARBolted beam-to-column connection may fail as aresult of block shear. Failure occurs in shear ata row of bolt holes parallel to the applied force,accompanied by tensile rupture along a perpen-dicular face. This type of failure results in a blockof material being torn out by the applied shearforce as shown in Fig. 4.56.

Block shear failure can be avoided by ensuringthat the applied shear force, Ft, does not exceedthe block shear capacity, Pr, given by

P

Fig. 4.55 Bracket bolted to column.

4.11.3.2 HSFG boltsIf parallel-shank or waisted-shank HSFG bolts, ratherthan ordinary bolts, were used in the connectiondetail shown in Fig. 4.50, failure of the connectionwould principally arise as a result of slip betweenthe connected parts. Thus, all connections utilisingfriction grip fasteners should be checked for slipresistance. However, connections using parallelshank HSFG bolts designed to be non-slip inservice, should additionally be checked for bearingcapacity of the connected parts and shear capacityof the bolts after slip.

Slip resistance. According to clause 6.4.2, theslip resistance of HSFG bolts designed to be non-slip in service, P sL, is given by

P sL = 1.1K sµPo (4.72)

and for HSFG bolts designed to be non-slip underfactored loads by

PsL = 0.9K sµPo (4.73)wherePo minimum shank tension (proof load)

(Table 4.24)K s = 1.0 for bolts in standard clearance holesµ slip factor ≤ 0.5


Table 4.24 Proof load of HSFG bolts, Po

Nominal size and thread Minimum shank tensiondiameter (mm) or proof load (kN)

12 49.416 92.120 14422 17724 20727 23430 28636 418

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224

Fig. 4.56 Block shear. (Based on Fig. 22, BS 5950)

Fr

Lv

Lt Lt

Fr

Lv

Pr = 0.6pyt[Lv + Ke(L t − kDt)] (4.77)

whereDt is the hole size for the tension facet is the thickness

Lt and L v are the dimensions shown in Fig. 4.56Ke is the effective net area coefficientk = 0.5 for a single line of bolts parallel to the

applied shear= 2.5 for two lines of bolts parallelto the applied shear

Example 4.21 Beam-to-column connection using web cleats(BS 5950)

Show that the double angle web cleat beam-to-column connection detail shown below is suitable to resist the designshear force, V, of 400 kN. Assume the steel is grade S275 and the bolts are M20, grade 8.8 in 2 mm clearance holes.

Bolt A1

356

× 36

8 ×

177

UC

gra

de S

275

50 4010

50

60

140

140

60

50

60

60

60

60

60

400

610 × 229 × 101 UBgrade S275

Angle cleats 90 × 90 × 10grade S275

50

CHECK FASTENER SPACING AND EDGE/END DISTANCESDiameter of bolt, d b = 20 mmDiameter of bolt hole, Dh = 22 mmPitch of bolt, p = 140 mm and 60 mmEdge distance, e1 = 40 mmEnd distance, e2 = 60 mm and 50 mmThickness of angle cleat, tp = 10 mm

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225


Example 4.21 continuedThe following conditions need to be met:

Pitch ≥ 2.5db = 2.5 × 20 = 50 < 140 and 60 OKPitch ≤ 14tp = 14 × 10 = 140 ≤ 140 and 60 OKEdge distance e1 ≥ 1.4Dh = 1.4 × 22 = 30.8 < 40 OKEnd distance e2 ≥ 1.4Dh = 1.4 × 22 = 30.8 < 60 and 50 OKe1 and e2 ≤ 11tpε = 11 × 10 × 1 = 110 < 40, 50 and 60 OK

(For grade S275 steel with tp = 10 mm, py = 275 N/mm2, ε = 1.) Hence all fastener spacing and edge/end distancesto fasteners are satisfactory.

CHECK STRENGTH OF BOLTS CONNECTING CLEATS TO SUPPORTING COLUMN

Shear6 No., M20 grade 8.8 bolts. Hence A s = 245 mm2 (Table 4.22) and ps = 375 N/mm2 (Table 4.19). Shear capacity ofsingle bolt, Ps, is

Ps = ps A s = 375 × 245 = 91.9 × 10 = 91.9 kN

Shear capacity of bolt group is

6Ps = 6 × 91.9 = 551.4 kN > V = 400 kN

Hence bolts are adequate in shear.

BearingBearing capacity of bolt, Pbb, is given by

Pbb = dbtpbb = 20 × 10 × 1000 = 200 × 103 = 200 kN

Since thickness of angle cleat (= 10 mm) < thickness of column flange (= 23.8 mm), bearing capacity of cleat iscritical. Bearing capacity of cleat, Pbs, is given by

Pbs = kbsdbtpbs = 1 × 20 × l0 × 460 = 92 × 103 N = 92 kN

≤ 0.5kbsetpbs = 0.5 × 1 × 60 × 10 × 460 = 138 × 103 N = 138 kN

Bearing capacity of connection is

6 × 92 = 552 kN > V = 400 kN

Therefore bolts are adequate in bearing.

CHECK STRENGTH OF BOLT GROUP CONNECTING CLEATS TO WEB OF SUPPORTED BEAMShear6 No., M20 grade 8.8 bolts; from above, As = 245 mm2 and ps = 375 N/mm2. Since bolts are in double shear, shearcapacity of each bolt is 2Ps = 2 × 91.9 = 183.8 kN

Loads applied to the bolt group are vertical shear, V = 400 kN and moment, M = 400 × 50 × 10−3 = 20 kN m.Outermost bolt (A l ) subject to greatest shear force which is equal to the resultant of the load due to the moment,

M = 20 kN m and vertical shear force, V = 400 kN. Load on the outermost bolt due to moment, Fmb, is given by

F

MZ

AA

Amb kN

. = =×

=20 10

42047 6

3

where A is the area of bolt and Z the modulus of the bolt group given by

Iy

A A

= =63 000

150420 mm3

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226

in which I is the inertia of the bolt group equal to

2A(302 + 902 + 1502) = 63000A mm4

Load on outermost bolt due to shear, Fvb, is given by

Fvb = V/No. of bolts = 400/6 = 66.7 kN

Resultant shear force of bolt, Fs, is

Fs = (F 2vb + F 2

mb)1/2 = (66.72 + 47.62)1/2 = 82 kN

Since Fs (= 82 kN) < 2Ps (= 183.8 kN) the bolts are adequate in shear.

BearingBearing capacity of bolt, Pbb, is

Pbb = 200 kN (from above) > Fs OK

Bearing capacity of each cleat, Pbs, is

Pbs = 92 kN (from above)

Bearing capacity of both cleats is

2 × 92 = 184 kN > Fs OK

Bearing capacity of the web, Pbs, is

Pbs = kbsdbtwpbs = 1 × 20 × 10.6 × 460 × 10−3 = 97.52 kN > Fs OK

Hence bolts, cleats and beam web are adequate in bearing.

SHEAR STRENGTH OF CLEATSShear capacity of a single angle cleat, Pv, is

Pv = 0.6py A v = 0.6 × 275 × 3600 × 10−3 = 594 kN

where

Av = 0.9Ao (clause 4.2.3 of BS 5950) = 0.9 × thickness of cleat (tp) × length of cleat (bp)

= 0.9 × 10 × 400 = 3600 mm2.

Since shear force V/2 (= 200 kN) < Pv (= 594 kN) the angle is adequate in shear.

BENDING STRENGTH OF CLEATS

M

V = × × =−

250 10

4002

3 × 50 × 10−3 = 10 kN m

Assume moment capacity of one angle of cleat, Mc, is

Mc = pyZ = 275 × 266.7 × 103 = 73.3 × 106 N mm = 73.3 kN m > M

where Z

t

= =

×p p2

6b 10 400

6

2

= 266 667 mm3. Angle cleat is adequate in bending.

LOCAL SHEAR STRENGTH OF THE BEAMShear capacity of the supported beam, Pv, is

Pv = 0.6p y Av = 0.6 × 275 × 6383.3 = 1053.2 × 103 N = 1053.2 kN > V (= 400 kN)

where A v = tw D = 10.6 × 602.2 = 6383.3 mm2. Hence supported beam at the end is adequate in shear.


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227

Example 4.22 Analysis of a bracket-to-column connection (BS 5950)Show that the bolts in the bracket-to-column connection below are suitable to resist the design shear force of200 kN. Assume the bolts are all M16, grade 8.8.


610

× 30

5 ×

179

UB

457 × 191 × 82 UB

250 P = 200 kN

100

100

100

70

Since the bolts are subject to combined shear and tension, the bolts should be checked for shear, tension andcombined shear and tension separately.

SHEARDesign shear force, P = 200 kNNumber of bolts, N = 8Shear force/bolt, Fs = P/N = 200/8 = 25 kN

Shear capacity of bolt, Ps, is

Ps = ps A s = 375 × 157 = 58.9 × 103 = 58.9 kN > Fs OK

TENSILE CAPACITYMaximum bolt tension, Ft, is

Ft = Pey1/2∑y 2

= 200 × 250 × 370/2(702 + 1702 + 2702 + 3702)

= 38 kN

Tension capacity, Pnom, is

Pnom = 0.8p t A t = 0.8 × 560 × 157 = 70.3 × 103 N

= 70.3 kN > Ft OK

COMBINED SHEAR AND TENSIONCombined check:

FP

FP

s

s

t

nom .+ ≤ 1 4

2558 9

3870 3

0 96 1 4.

.

. .+ = ≤ OK

Hence the M16, grade 8.8 bolts are satisfactory.

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228

Example 4.23 Analysis of a beam splice connection (BS 5950)Show that the splice connection shown below is suitable to resist a design bending moment, M, and shear force, F,of 270 kN m and 300 kN respectively. Assume the steel grade is S275 and the bolts are general grade, M22, parallel-shank HSFG bolts. The slip factor, µ, can be taken as 0.5.

457 × 191 × 89 UB

100

40

100

100

40

40 90

170Web coverplates

(2)10 mm thickweb coverplates

40 80 80 80 90 80 80 80 40

Flange cover plate180 × 15 × 650

90

650

40

380

Assume that (1) flange cover plates resist the design bending moment, M and (2) web cover plates resist the designshear force, F, and the torsional moment (= Fe), where e is half the distance between the centroids of the bolt groupseither side of the joint.

FLANGE SPLICECheck bolts in flange cover plateSingle cover plate on each flange. Hence the bolts in the flange are subject to single shear and must resist a shearforce equal to

Applied momentOverall depth of beam

kN

. . = =

×=

MD

270 10463 6

582 43

Slip resistance of single bolt in single shear, PsL, is

PsL = 1.1K sµPo = 1.1 × 1.0 × 0.5 × 177 = 97.3 kN (Po = 177 kN, Table 4.24)

Since thickness of beam flange (= 17.7 mm) > thickness of cover plate (= 15 mm), bearing in cover plate is critical.Bearing capacity, Pbg, of cover plate is

1.5dbtpbs ≤ 0.5etpbs

1.5dbtpbs = 1.5 × 22 × 15 × 460 = 227.7 × 103 N ( pbs = 460 N/mm2, Table 4.21)

0.5etpbs = 0.5 × 40 × 15 × 460 = 138 × 103 N

Shear capacity of the bolts after slip, Ps, is

Ps = psAs = 400 × 303 × 10−3 = 121.2 kN per bolt

where As = At = 303 mm2 (Table 4.22)

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229

Hence shear strength based on slip resistance of bolts. Slip resistance of 8 No., M22 bolts = 8PsL = 8 × 97.3 =778.4 kN > shear force in bolts = 582.4 kN. Therefore the 8 No., M22 HSFG bolts provided are adequate in shear.

Check tension capacity of flange cover plateGross area of plate, αg, is

αg = 180 × 15 = 2700 mm2

Net area of plate, αn, is

αn = αg − area of bolt holes

= 2700 − 2(15 × 24) = 1980 mm2

Force in cover plate, Ft, is

F

MD Tt

fp

kN

.

. =+

=×

+=

270 10463 6 15

564 13

where Tfp is the thickness of flange cover plate = 15 mm.Tension capacity of cover plate, Pt, is

Pt = α epy = 2376 × 275 = 653.4 × 103 N = 653.4 kN

where αe = K eαn = 1.2 × 1980 = 2376 mm2.Since Pt > Ft, cover plate is OK in tension.

Check compressive capacity of flange cover plateTop cover plate in compression will also be adequate provided the compression flange has sufficient lateral restraint.

WEB SPLICECheck bolts in web spliceVertical shear Fv = 300 kN

Torsional moment = Fv eo (where eo is half the distance between the centroids of the bolt groups either side of thejoint in accordance with assumption (2) above)

= 300 × 45 = 13 500 kN mm

Maximum resultant force FR occurs on outermost bolts, e.g. bolt A, and is given by the following expression:

FR = F Fvs2

tm2 +

where Fvs is the force due to vertical shear and Ftm the force due to torsional moment.

F

FNvs

v kN = = =3004

75

F

A

Ztmb

b

Torsional moment =

×

whereAb area of single boltZb modulus of the bolt group which is given by

Inertia of bolt groupDistance of furthest bolt

2 bb

( ) .=

+=

AA

50 150150

333 332 2

Hence

F

AAtm

b

b333.33kN

. = =

13 50040 5



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230

and FR = 40 5 752 2. + = 85.2 kN

From above, slip resistance of M22 HSFG bolt, PsL = 97.3 kN. Since two cover plates are present, slip resistance perbolt is

2PsL = 2 × 97.3 = 194.6 kN > FR

Therefore, slip resistance of bolts in web splice is adequate.Similarly, shear capacity of bolt in double shear, Ps = 2 × 121.2 = 242.4 kN > FR

Therefore, shear capacity of bolts in web splice after slip is also adequate.

Check web of beamThe forces on the edge of the holes in the web may give rise to bearing failure and must therefore be checked. Heree = edge distance for web and

tan .θ θ= ⇒ = °

40.575

28 37

e

. .= = =

45 450 475

94 7sin θ

100

100

100

Bolt AFtm

Fvs θ FR

θe

45

e0

Bearing capacity of web, Pbg, is

= 1.5dbtpbs ≤ 0.5etp bs = 0.5 × 94.7 × 10.6 × 460 = 230.9 × 103 N = 203.9 kN

= 1.5 × 22 × 10.6 × 460 = 160.9 × 103 N = 160.9 kN > FR

Hence web of beam is adequate in bearing.

Check web cover platesThe force on the edge of the holes in the web cover plates may give rise to bearing failure and must therefore bechecked. Here e = edge distance for web cover plates and θ = 28.37°. Thus

e

. .= = =

40cos θ

400 88

45 5

Bearing capacity of one plate is

= 1.5dbtpbs ≤ 0.5etp bs = 0.5 × 45.5 × 10 × 460 = 104.7 × 103 N = 104.7 kN= 1.5 × 22 × 10 × 460 = 151.8 × 103 N = 151.8 kN


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231



100

100

100

40

380

170

40 4090

Ftm

FR Fvs

e

40θ

Thus bearing capacity of single web plate = 104.7 kN and bearing capacity of two web plates = 2 × 104.7 = 209 kN> FR. Hence web cover plates are adequate in bearing.

Check web cover plates for shear and bendingCheck shear capacity. Shear force applied to one plate is

Fv/2 = 300/2 = 150 kN

Torsional moment applied to one plate is

150 × 45 = 6750 kN mm = 6.75 kN m

24 mm

Dh

Dh

50

150

380

Shear capacity of one plate, Pv, is

Pv = 0.6py Av = 0.6 × 275 × 0.9 × 10(380 − 4 × 24) = 421.7 kN > 150 kN

where A v = 0.9An. Hence cover plate is adequate in shear.

Check moment capacitySince applied shear force (= 150 kN) < 0.6Pv = 253 kN, moment capacity, Mc = pyZ where Z = I/y.

Referring to the above diagram, I of plate = 10 × 3803/12 − 2(10 × 24)502 − 2(10 × 24)1502 = 337 × 105 mm2

Z = I/y = 337 × 105/190 = 177 × 103 mm2

Mc = pyZ = 275 × 177 × 103 = 48.7 × 106 N mm = 48.7 kN m > 6.75 kN m

Hence web cover plates are also adequate for bending.

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232

Example 4.24 Analysis of a beam-to-column connection using anend plate (BS 5950)

Calculate the design shear resistance of the connection shown below, assuming that the steel is grade S275 and thebolts are M20, grade 8.8 in 2 mm clearance holes.

6 mm filletweld

610 × 229 × 101 UB

Grade 8.8, M20 bolts

tp = 10 mm356

× 36

8 ×

177

UC

35

50

35

120120120

50

50120120120

50

e1

e2

pp

35 90 35 Pvp Pvp

�p = 460 mm

CHECK FASTENER SPACING AND EDGE/END DISTANCES

Diameter of bolt, d b = 20 mmDiameter of bolt hole, Dh = 22 mmPitch of bolt, p = 120 mmEdge distance, e1 = 35 mmEnd distance, e2 = 50 mmThickness of end plate, tp = 10 mm

The following conditions need to be met:

Pitch ≥ 2.5db = 2.5 × 20 = 50 < 120 OKPitch ≤ 14tp = 14 × 10 = 140 > 120 OKEdge distance e1 ≥ 1.4Dh = 1.4 × 22 = 30.8 < 35 OKEnd distance e2 ≥ 1.4Dh = 1.4 × 22 = 30.8 < 50 OKe1 and e2 ≤ 11tpε = 11 × 10 × 1 = 110 > 35, 50 OK

For grade S275 steel with tp = 10 mm, py = 275 N/mm2 (Table 4.3), ε = 1 (equation 4.4). Hence all fastener spacingand edge/end distances to fasteners are satisfactory.

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233


BOLT GROUP STRENGTH

Shear8 No., M20 grade 8.8 bolts; A s = 245 N/mm2 (Table 4.22) and ps = 375 N/mm2 (Table 4.19)

Shear capacity of single bolt, Ps, is

Ps = psA s = 375 × 245 = 91.9 × 103 = 91.9 kN

Shear capacity of bolt group is

8Ps = 8 × 91.9 = 735 kN

BearingBearing capacity of bolt, Pbb, is given by

Pbb = dbtppbb = 20 × 10 × 1000 = 200 × 103 = 200 kN

End plate is thinner than column flange and will therefore be critical. Bearing capacity of end plate, Pbs, is given by

Pbs = kbsdbtppbs = 1 × 20 × 10 × 460 = 92 × 103 N = 92 kN

≤ 0.5kbse2tppbs = 0.5 × 1 × 50 × 10 × 460 = 115 × 103 N = 115 kN

Hence bearing capacity of connection = 8 × 92 = 736 kN.

END PLATE SHEAR STRENGTH

Av = 0.9An = 0.9tp(bp − 4Dh) = 0.9 × 10 × (460 − 4 × 22) = 3348 mm2

Shear capacity assuming single plane of failure, Pvp, is

Pvp = 0.6py A v = 0.6 × 275 × 3348 = 552.4 × 103 = 552.4 kN

Shear capacity assuming two failure planes is

2Pvp = 2 × 552.4 = 1104.8 kN

WELD STRENGTH(Readers should refer to section 4.11.5 before performing this check.) Fillet weld of 6 mm provided. Hence

Leg length, s = 6 mm

Design strength, pw = 220 N/mm2 (assuming electrode strength = 42 N/mm2, Table 4.25 )Throat size, a = 0.7s = 0.7 × 6 = 4.2 mmEffective length of weld, bw, is

bw = 2(bp − 2s) = 2(460 − 2 × 6) = 896 mm

Hence design shear strength of weld, Vw, is

Vw = pwabw = 220 × 4.2 × 896 = 828 × 103 N = 828 kN

LOCAL SHEAR STRENGTH OF BEAM WEB AT THE END PLATE

Pvb = 0.6py A v = 0.6 × 275(0.9 × 10.6 × 460) × 10−3 = 724 kN

Hence, strength of connection is controlled by shear strength of beam web and is equal to 724 kN.


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234

Size orleg length

Throat(b)

(a)

Throat

Fig. 4.57 Types of weld; (a) butt weld; ( b) fillet weld.

4.11.5 WELDED CONNECTIONSThe two main types of welded joints are fillet weldsand butt welds. Varieties of each type are shownin Fig. 4.57. Essentially the process of welding con-sists of heating and melting steel in and/or aroundthe gap between the pieces of steel that are beingwelded together. Welding rods consist of a steelrod surrounded by a flux which helps the metal tomelt and flow into the joint. Welding can be ac-complished using oxy-acetylene equipment, but theeasiest method uses electric arc welding.

4.11.5.1 Strength of weldsIf welding is expertly carried out using the correctgrade of welding rod, the resulting weld should beconsiderably stronger than the pieces held together.However, to allow for some variation in the qualityof welds, it is assumed that the weld strength forfillet welds is as given in Table 37 of BS 5950,reproduced below as Table 4.25.

However, the design strength of the weld can betaken as the same as that for the parent metal if thejoint is a butt weld, or alternatively a fillet weldsatisfying the following conditions:

1. The weld is symmetrical as shown in Fig.4.58.

2. It is made with suitable electrodes which willproduce specimens at least as strong as the par-ent metal.

3. The sum of throat sizes (Fig. 4.58) is not lessthan the connected plate thickness.

4. The weld is principally subject to direct tensionor compression (Fig. 4.58).

4.11.5.2 Design detailsFigure 4.57 also indicates what is meant by theweld leg length, s, and the effective throat size,a, which should not be taken as greater than0.7s.

Fig. 4.58 Special fillet weld.

Table 4.25 Design strength of fillet welds, pw

(Table 37, BS 5950)

Steel grade Electrode classification

35 42 50

N/mm2 N/mm2 N/mm2

S275 220 220a 220a

S355 220b 250 250a

S460 220b 250b 280

Notes. a Over matching electrodes.b Under matching electrodes. Not to be used for partialpenetration butt welds.

t

Throat a

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235


The effective length of a run of weld should betaken as the actual length, less one leg length foreach end of the weld. Where the weld ends at a cornerof the metal, it should be continued around thecorner for a distance greater than 2s. In a lap joint,

the minimum lap length should be not less than 4t,where t is the thinner of the pieces to be joined.

For fillet welds, the ‘vector sum of the designstresses due to all forces and moments transmittedby the weld should not exceed the design strength, pw’.

Example 4.25 Analysis of a welded beam-to-column connection(BS 5950)

A grade S275 steel 610 × 229 × 101 UB is to be connected, via a welded end plate onto a 356 × 368 × 177 UC. Theconnection is to be designed to transmit a bending moment of 500 kN m and a shear force of 300 kN. Show that theproposed welding scheme for this connection is adequate.

Beam

15 mmend plate

Elevation

Col

umn

356

× 36

8 ×

177

UC

610 × 229 × 101 UB

15 mm stiffeners

Plan End view

Electrode strength42

10 mm fillet welds

10 mm fillet welds

225

y

y1

x x

y

570

587

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236

4.12 SummaryThis chapter has considered the design of a numberof structural steelwork and composite elements,including beams, slabs, columns and connections,to BS 5950: Structural use of steel work in buildings.The ultimate limit state of strength and theserviceability limit state of deflection principallyinfluence the design of steel elements. Many steelstructures are still analaysed by assuming thatindividual elements are simply supported at theirends. Steel sections are classified as plastic, compact,semi-compact or slender depending on how thesections perform in bending. The design of flexuralmembers generally involves considering the limitstates of bending, shear, web bearing/buckling,deflection and if the compression flange is not fully

restrained, lateral torsional buckling. Columns sub-ject to axial load and bending are normally checkedfor cross-section capacity and buckling resistance.

The two principal methods of connecting steelelements of a structure are bolting and welding. Itis vitally important that the joints are designed toact in accordance with the assumptions made inthe design. Design of bolted connections, usingordinary (or black) bolts, usually involves check-ing that neither the bolt nor the elements beingjoined exceed their shear, bearing or tensioncapacities. Where HSFG bolts are used, the slipresistance must also be determined. Welded con-nections are most often used to weld end platesor cleats to members, a task which is normallyperformed in the fabrication yard rather than onsite.

Example 4.25 continuedLeg length of weld, s = 10 mm. Effective length of weld is

4(bw − 2s) + 2(hw − 2s) = 4(225 − 2 × 10) + 2(570 − 2 × 10) = 1920 mm

Weld force

shear forceeffective length of weld

kN/mm .= = =300

19200 16

Weld second moment of area, Ixx, is

Ixx = 4[1 × (Bw − 2s)]y21 + 2[1 × (hw − 2s)3/12]

= + × [( )] . [ ( ) / ]4 205 293 5 21 550 122 3

= 98 365 812 mm4

Weld shear (moment) is

MyIxx

kN/mm .

. =× ×

=500 10 293 5

98 365 8121 49

3

where y = D/2 = 587/2 = 293.5 mmVectored force = 1 49 0 162 2. .+ = 1.50 kN/mmSince the steel grade is S275 and the electrode strength is 42, the design strength of the weld is 220 N/mm2

(Table 4.25).Weld capacity of 10 mm fillet weld, pwc, is

pwc = apw = 0.7 × 10 × 220 × 10−3 = 1.54 kN/mm > 1.50 kN/mm OK

where throat thickness of weld, a = 0.7s = 0.7 × 10 mmHence proposed welding scheme is just adequate.

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237

Summary

Questions1. A simply-supported beam spanning 8 m

has central point dead and imposed loadsof 200 kN and 100 kN respectively.Assuming the beam is fully laterallyrestrained select and check suitableUB sections in (a) S275 and (b) S460steel.

2. A simply-supported beam spanning 8 mhas uniformly distributed dead andimposed loads of 20 kN/m and 10 kN/mrespectively. Assuming the beam is fullylaterally restrained select and check suitableuniversal beam sections in (a) grade S275and (b) grade S460 steel.

40 kN/m dead load

12 kN imposed load

+ 32 kN/m imposed load

12 kN imposed load

8000 2000

Fig. 4.59

3. For the fully laterally restrained beamshown in Fig. 4.59 select and check asuitable UB section in grade S275 steelto satisfy shear, bending and deflectioncriteria.

4. If the above beam is laterally unrestrained,select and check a suitable section ingrade S275 steel to additionally satisfylateral torsional buckling, web bearingand buckling criteria. Assume that eachsupport is 50 mm long and lateralrestraint conditions at supports are asfollows: –

Compression flange laterally restrained.Beam fully restrained against torsion.Both flanges free to rotate on plan.Destabilising load conditions.

5. If two discrete lateral restraints, one atmid-span, and one at the cantilever tip,are used to stabilise the above beam,select and check a suitable section ingrade S275 steel to satisfy all the criteriain question 4.

6. Carry out designs for the beams shown inFig. 4.60.

7. (a) List and discuss the merits of floorsystems used in steel framed structures.

(b) A floor consists of a series of beams8.0 m span and 4.0 m apart,supporting a reinforced concrete slab120 mm thick. The superimposedload is 4 kN/m2 and the weight offinishes is 1.2 kN/m2.

(i) Assuming the slab and supportbeams are not connected, selectand check a suitable UB sectionin grade S355 steel.

(ii) Repeat the above design assumingthe floor is of compositeconstruction. Comment on yourresults. Assume the unit weight ofreinforced concrete is 24 kN/m3.

8. (a) List and discuss the factors thatinfluence the load carrying capacityof steel columns.

(b) A 254 × 254 × 132 universal columnin grade S275 steel is required tosupport an ultimate axial compressiveload of 1400 kN and a major axismoment of 160 kN m applied at thetop of the element. Assuming thecolumn is 5 m long and effectivelypin ended about both axes, check thesuitability of the section.

9. A 305 × 305 × 137 UC section extendsthrough a height of 3.5 m.(a) Check that the section is suitable for

plastic design using grade S275 andgrade S355 steel.

(b) Calculate the squash load for gradeS275 steel.

(c) Calculate the full plastic moment ofresistance for grade S275 steel aboutthe major axis.

(d) Find the reduced plastic moment ofresistance about the major axis whenF = 800 kN using grade S275 steel.

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238

50 kNimposed load

20 kN m–1

dead load

3 m

50 kN 50 kN 50 kN 50 kNimposed loads

3 m

30 kNimposed load

30 kNimposed load

10 kN m–1 dead load

5 m 2.5 m

100 kNimposed load

20 kN m–1

dead load20 kN m–1

dead load

2 m 4 m 2 m

100 kNimposed load

Fig. 4.60

250 mm500 kN

350 mm

175 mm

10 mm fillet welds

ss

12. Design a splice connection for a 686 ×254 × 140 UB section in grade S275 steelto cater for half bending strength andhalf the shear capacity of the section.

13. (a) Explain with the aid of sketches thefollowing terms used in weldedconnections:(i) fillet welds(ii) butt welds(iii) throat size.

(b) A bracket made from a 406 × 178 ×60 universal beam is welded to asteel column as shown below. Thebracket is designed to support anultimate load of 500 kN. Show theproposed welding scheme for thisconnection is adequate. Assume thesteel is of grade S275 and theelectrode strength is 42.

(e) When Mx = 500 kN, check thelocal and overall capacity of thecolumn assuming the base is pinned.

10. Select a suitable short column section ingrade S275 steel to support a factoredaxial concentric load of 1000 kN andfactored bending moments of 400 kN mabout the major axis, and 100 kN mabout the minor axis.

11. Select a suitable column section in gradeS275 steel to support a factored axialconcentric load of 1000 kN and factoredbending moments of 400 kN m aboutthe major axis, and 100 kN m about theminor axis, both applied at each end.The column is 10 m long and is fullyfixed against rotation at top and bottom,and the floors it supports are bracedagainst sway.

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Chapter 5

Design inunreinforced masonryto BS 5628

concrete due largely to the research and marketingwork sponsored in particular by the Brick Devel-opment Association. For instance, everybody nowknows that ‘brick is beautiful’. Less well appreciated,

Fig. 5.1 Traditional application of masonry in construction: (a) brick bridge; (b) brick sewer; (c) brick retaining wall.

This chapter is concerned with the design of unreinforcedmasonry walls to BS 5628: Part 1. The chapter describesthe composition and properties of the three main mate-rials used in masonry construction: bricks, blocks andmortars. Masonry is primarily used nowadays in theconstruction of load-bearing and non-load-bearing walls.The primary aim of this chapter is to give guidance onthe design of single leaf and cavity walls, with andwithout stiffening piers, subject to either vertical orlateral loading.

5.1 IntroductionStructural masonry was traditionally very widelyused in civil and structural works including tun-nels, bridges, retaining walls and sewerage systems(Fig. 5.1). However, the introduction of steel andconcrete with their superior strength and costcharacteristics led to a sharp decline in the use ofmasonry for these applications.

Over the past two decades or so, masonry hasrecaptured some of the market lost to steel and

(a) (c)

(b)

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Design in unreinforced masonry to BS 5628

240

Part 3: Materials and Components, Design andWorkmanship

Part 1 was originally published in 1978 andParts 2 and 3 were issued in 1985. This book onlydeals with the design of unreinforced masonrywalls, subject to vertical or lateral loading. It should,therefore, be assumed that all references to BS 5628refer to Part 1, unless otherwise noted.

Before looking in detail at the design of masonrywalls, the composition and properties of the com-ponent materials are considered in the followingsections.

5.2 MaterialsStructural masonry basically consists of bricks orblocks bonded together using mortar or grout.In cavity walls, wall ties complying with BS EN845-1 are also used to tie together the two skinsof masonry. Fig. 5.3 shows some examples of thewall ties used in masonry construction. In externalwalls, and indeed some internal walls dependingon the type of construction, a damp proof courseis also necessary to prevent moisture ingress to thebuilding fabric (Fig. 5.4).

The following discussion will concentrate onthe composition and properties of the three maincomponents of structural masonry, namely (i)bricks, (ii) blocks and (iii) mortar. The reader isreferred to BS 5628: Parts 1 and 3 and BS EN845-1 for guidance on the design and specificationof wall ties and damp-proof systems for normalapplications.

perhaps, is the fact that masonry has excellent struc-tural, thermal and acoustic properties. Furthermore,it displays good resistance to fire and the weather.There are undisputed advantages of using brick-work masonry in flood prone areas. On this basisit has been argued that it is faster and cheaper tobuild certain types of buildings using only masonryrather than using a combination of materials toprovide these properties.

Brick manufacturers have also pointed out that,unlike steelwork, masonry does not require regularmaintenance nor, indeed, does it suffer from thedurability problems which have plagued concrete.The application of reinforcing and prestressing tech-niques to masonry has considerably improved itsstructural properties and hence the appeal of thismaterial to designers.

Despite the above, masonry is primarily usednowadays for the construction of load-bearing andnon-load-bearing walls (Fig. 5.2). These structuresare principally designed to resist lateral and vert-ical loading. The lateral loading arises mainly fromthe wind pressure acting on the wall. The verticalloading is attributable to dead plus imposed load-ing from any supported floors, roofs, etc. and/orself-weight of the wall.

Design of masonry structures in the UK is car-ried out in accordance with the recommendationsgiven in BS 5628: Code of Practice for Use of Masonry.This code is divided into three parts:

Part 1: Structural Use of Unreinforced MasonryPart 2: Structural Use of Reinforced and Prestressed

Masonry

Fig. 5.2 (a) Load-bearing and (b) non-load-bearing walls.

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241

5.2.1 BRICKSBricks are manufactured from a variety of materialssuch as clay, calcium silicate (lime and sand/flint),concrete and natural stone. Of these, clay bricksare by far the most commonly used variety in theUK.

Clay bricks are manufactured by shaping suit-able clays to units of standard size, normally takento be 215 × 102.5 × 65 mm (Fig. 5.5). Sand facingsand face textures may then be applied to the ‘green’clay. Alternatively, the clay units may be perforatedor frogged in order to reduce the self-weight of theunit. Thereafter, the clay units are fired in kilns toa temperature in the range 900–1500 °C in orderto produce a brick suitable for structural use. Thefiring process significantly increases both thestrength and durability of the units.

In design it is normal to refer to the coordinatingsize of bricks. This is usually taken to be 225 ×112.5 × 75 mm and is based on the actual or work

Fig. 5.3 Wall ties to BS EN 845-1: (a) butterfly tie; (b) double triangle tie; (c) vertical twist tie.

Fig. 5.6 Coordinating and work size of bricks.

size of the brick, i.e. 215 × 102.5 × 65 mm, plus anallowance of 10 mm for the mortar joint (Fig. 5.6).Clay bricks are also manufactured in metric modu-lar format having a coordinating size of 200 × 100× 75 mm. Other cuboid and special shapes are alsoavailable (BS 4729).

Fig. 5.4 Damp proof courses: (a) lead, copper, polythene,bitumen polymer, mastic asphalt; (b) two courses of bondedslate; (c) two courses of d.p.c. bricks (based on Table 2,BS 5628: Part 3).

Fig. 5.5 Types of bricks: (a) solid; (b) perforated;(c) frogged.

Materials

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According to BS EN 771-1, clay bricks canprimarily be specified in terms of:

• density• compressive strength• durability• active soluble salt content• water absorption

Clay bricks with a gross density of less than orequal to 1000 kg/mm3 are classified as LD (lowdensity) units and those with a gross density exceed-ing 1000 kg/m3 as HD (high density) units. Claybricks available in the UK are generally HD typeunits.

The compressive strength of bricks is a func-tion of the raw materials and the actual processesused to manufacture the units. Bricks are generallyavailable with compressive strengths ranging from5 N/mm2 to more than 150 N/mm2.

BS EN 771-1 defines three durability categoriesfor clay bricks, namely F0, F1 and F2, whichindicate their susceptibility to frost damage insaturated or near saturated conditions (Table 5.1).For example, external brickwork near ground levelwith a high risk of saturation, poorly drained withfreezing, would typically be built using category F2brick. On the other hand, internal walls could bemade using category F0 bricks. There are also threecategories of soluble salt content defined in thestandard (Table 5.2). A high soluble salt contentincreases the risk of sulphate attack of the mortarand, possibly, of the bricks themselves, if there is aconsiderable amount of water movement through

the masonry. A more common problem associatedwith soluble salts is that of efflorescence stainingof brickwork. This is a white deposit caused bythe crystallisation of soluble salts in brickwork as aresult of wet masonry drying out. Efflorescence isnot structurally significant but can mar brickwork’sappearance. Magnesium sulphate may causedamage to the units and it is for this reason themagnesium content is specified separately.

Brick manufacturers are also required to declarethe water absorption category (expressed as apercentage increase in weight) of the unit as thisinfluences flexural strength (see section 5.6). Thewater absorption categories for which flexuralstrength data is available in the UK are: less than7 per cent, 7–12 per cent and > 12 per cent andare determined using the test procedure describedin Annex C of BS EN 771-1.

Clay bricks are often referred to as commons,facing, engineering and DPC, which reflects theirusage. Common bricks are those suitable for gen-eral construction work, with no special claim togive an attractive appearance. Facing bricks on theother hand are specially made or selected to givean attractive appearance on the basis of colour andtexture. Engineering bricks tend to be high densityand are designated class A and class B on the basisof strength, water absorption, frost resistance andthe maximum soluble salt content (Table 5.3). Claydamp-proof course bricks (DPC 1 and DPC 2) areclassified in terms of their water absorption propertyonly.

Guidance on the selection of clay bricks mostappropriate for particular applications can be foundin Table 12 of BS 5628: Part 3 (see below).

5.2.2 BLOCKSBlocks are walling units but, unlike bricks, arenormally made from concrete. They are availablein two basic types: autoclaved aerated concrete (nowreferred to as aircrete) and aggregate concrete.The aircrete blocks are made from a mixture ofsand, pulverised fuel ash, cement and aluminiumpowder. The latter is used to generate hydrogenbubbles in the mix; none of the powder remainsafter the reaction. The aggregate blocks have a com-position similar to that of normal concrete, con-sisting chiefly of sand, coarse and fine aggregateand cement plus extenders. Aircrete blocks tendto have lower densities (typically 400–900 kg/m3)than aggregate blocks (typically 1200–2400 kg/m3)which accounts for the former’s superior thermalproperties, lower unit weight and lower strengths.Commonly produced compressive strength of

Table 5.2 Soluble salt categories for clay bricks(Table 1, EN 771-1)

Category Total % by mass not greater than

Na+ + K+ Mg 2+

S0 No requirements No requirementsS1 0.17 0.08S2 0.06 0.03

Table 5.1 Durability categories for clay bricks

Durability category Frost resistance

F0 Passive exposureF1 Moderate exposureF2 Severe exposure

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Web

Shell

Table 5.3 Classification of clay engineering and DPC bricks (based on Table NA.6, BS EN 771-1)

Type Declared average Average absorption Freeze/thaw Solublecompressive strength (% by weight) not resistance sulphatenot less than (N/mm2) greater than

Engineering A 125 4.5 F2 S2Engineering B 75 7.0 F2 S2

Damp-proof course 1 – 4.5 – –Damp-proof course 2 – 7.0 – –

Materials

Table 5.5 Compressive strength of concreteblocks (BS EN 771-3)

Compressive strengths (N/mm2)

2.93.65.27.3

10.417.522.530.040.0

aircrete blocks are 2.9, 3.6, 4.2, 7.3 and 8.7N/mm2. Generally, aircrete blocks are more expen-sive than aggregate blocks.

Blocks are manufactured in three basic forms:solid, cellular and hollow (Fig. 5.7). Solid blockshave no formed holes or cavities other than thoseinherent in the material. Cellular blocks have oneor more formed voids or cavities which do not passthrough the block. Hollow blocks are similar tocellular blocks except that the voids or cavities passthrough the block. As discussed in section 5.5.2.1,the percentage of formed voids in blocks (andformed voids or frogs in bricks) influences the char-acteristic compressive strength of masonry.

Fig. 5.7 Concrete blocks: (a) solid; (b) cellular;(c) hollow.

For structural design, the two most importantproperties of blocks are their size and compressivestrength. Tables 5.4 and 5.5 give the most com-monly available sizes and compressive strengths ofconcrete blocks in the UK. The most frequentlyused block has a work face of 440 × 215 mm,width 100 mm and compressive strength 3.6 N/mm2;2.9 N/mm2 is a popular strength for aircrete blocks,and 7.3 N/mm2 for aggregate concrete blocks asit can be used below ground. Guidance on theselection and specification of concrete blocks inmasonry construction can be found in BS 5628:Part 3 and BS EN 771-3 respectively.

Table 5.4 Work sizes of concrete blocks (Table NA.1,BS EN 771-3)

Length Height Width (mm)(mm) (mm)

75 90 100 140 150 190 200 215 225

390 190 – x x x – x x – –440 215 x x x x x x x x x

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244

Table 5.7 Desirable minimum quality of brick, blocks and mortar for durability (based on Table 12,BS 5628: Part 3)

Location or type of wall Clay bricks Concrete blocks Mortardesignations

NF a FH a

Internal non-load-bearing walls NFa (F0)b All types, t ≥ 75 mm (iv) (iii)FHa (F1)b

Internal load-bearing walls, inner leaf of NFa (F0)b All types, t ≥ 90 mm (iv) (iii)cavity walls FHa (F1)b Sd

External walls; outer leaf of cavity walls ADc (F1)b All types, t ≥ 90 mm (iv)e (iii)f

BDc (F2)b Sd (iii)g (iii)g

External freestanding walls (with coping) F2b Sd (iii) (iii)Earth retaining wall (back filled with F2b Sd (ii)g,h (ii)g,h

free draining material)

Notes:aNF, no risk of frost during construction; FH, frost hazard a possibility during constructionbF0, passive exposure; F1, moderate exposure; F2, severe exposurecAD, above damp proof course; BD, below damp proof coursedBlocks should comply with one of the following: net density ≥ 1500 kgm−3 or be made with dense aggregate or have acompressive strength ≥ 7.3 Nmm−2

eFor clay brickwork, mortar designation not less that (iii)fIf brickwork is to be rendered the use of plasticised mortars is desirablegWhere sulphates are present in the soil or ground water, use sulphate resisting Portland cementhFor clay bricks, mortar designation (i) should be used

Table 5.6 Types of mortars (Table 1, BS 5628)

Mortar Compressive Prescribed mortars (proportion of materials by volume) Compressivedesignation strength strength at

class Cement-lime- Cement-sand Masonry Masonry 28 dayssand with or with or cement1-sand cement2-sand (N/mm2)without air without airentrainment entrainment

Increasing (i) M12 1:0 to 1/4:3 – – – 12ability to (ii) M6 1:1/2:4 to 41/2 1:3 to 4 1:21/2 to 31/2 1:3 6accommodate (iii) M4 1:1:5 to 6 1:5 to 6 1:4 to 5 1:31/2 to 4 4movement, (iv) M2 1:2:8 to 9 1:7 to 8 1:51/2 to 61/2 1:41/2 2e.g. due tosettlement,temperatureand moisturechanges

Notes:1Masonry cement with organic filler other than lime2Masonry cement with lime

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Symbols

For most forms of masonry construction a 1:1:6cement-lime-sand mortar is suitable. A simpleroption may be to specify a masonry cement whichconsists of a mixture of approximately 75 per centOPC, an integral air entraining agent and fineinorganic fillers, e.g. calcium carbonate, or lime.Such mixes will have similar properties to thosedisplayed by cement-lime-sand plasticised mortars.

By varying the proportions of the constituentsreferred to above, mortars of differing compressiveand bond strength, plasticity and frost resistantproperties can be produced. Table 5.6 shows thecomposition of recommended types of mortars formasonry construction. Table 5.7, which is based onTable 12 of BS 5628: Part 3, gives guidance on theselection of masonry units and mortars for use indifferent applications.

5.3 Masonry designThe foregoing has summarised some of the moreimportant properties of the component materialswhich are used in masonry construction. As notedat the beginning of this chapter, masonry construc-tion is mostly used nowadays in the constructionof load-bearing and panel walls. Load-bearing wallsprimarily resist vertical loading whereas panel wallsprimarily resist lateral loading. Figure 5.8 showssome commonly used wall types.

The remainder of this chapter considers thedesign of (1) load-bearing walls, with and withoutstiffening piers, resisting vertical compression load-ing (section 5.5) and (2) panel walls resisting lateralloading (section 5.6).

5.4 SymbolsFor the purposes of this chapter, the following sym-bols have been used. These have largely been takenfrom BS 5628.

GEOMETRIC PROPERTIES:t actual thickness of wall or leafh height of panel between restraintsL length of wall between restraintsA cross-sectional area of loaded wallZ sectional modulustef effective thickness of wallhef effective height of wallK effective thickness coefficientex eccentricity of loading at top of wallSR slenderness ratioβ capacity reduction factor

5.2.3 MORTARSThe primary function of the mortar is to bind to-gether the individual brick or block units, therebyallowing allow the transfer of compression, shearand tensile stresses between adjacent units, as wellas provide an effective seal between masonry unitsagainst rain penetration. However, there are severalother properties which the mortar must possess forease of construction and maintenance. For instance,the mortar should be easy to spread and remainplastic for a sufficient length of time in order thatthe units can be accurately positioned before settingoccurs. On the other hand, the setting time shouldnot be too excessive otherwise the mortar may besqueezed out as successive courses of units are laid.The mortar should be able to resist water uptakeby the absorbent bricks/blocks, e.g. by incorporatinga water-retaining admixture and/or use of a mortartype that includes lime, otherwise hydration andhence full development of the mortar strength maybe prevented. The mortar must also be durable. Forexample, if masonry remains wet for extended periodsof time, the mortar may be susceptible to sulphateattack due to the presence of sulphate in claymasonry units, the groundwater or the soil. Masonrymortar is also susceptible to freeze/thaw attack,particularly when newly laid, which can adverselyaffect bond strength. It should be remembered toothat the appearance of mortar is also important andthat it should be in harmony with the masonry unit.

In its most basic form, mortar simply consists of amixture of sand and ordinary Portland cement(OPC). However, such a mix is generally unsuitablefor use in masonry (other than perhaps in founda-tions and below damp-proof courses) since it willtend to be too strong in comparison with the strengthof the bricks/blocks. It is generally desirable to pro-vide the lowest grade of mortar possible, taking intoaccount the strength and durability requirements ofthe finished works. This is to ensure that any crack-ing which occurs in the masonry due to settlementof the foundations, thermal/moisture movements,etc. will occur in the mortar rather than the masonryunits themselves. Cracks in the mortar tend to besmaller because the mortar is more flexible than themasonry units and, hence, are easier to repair.

In order to produce a more practicable mix, itis normal to add lime to cement-sand mortar. Thishas the effect of reducing the strength of the mix,but at the same time increasing its workability andbonding properties. However, the mortar becomesmore susceptible to frost attack in saturated, ornear saturated, conditions but this can be offsetby adding an air-entraining (plasticising) agent.

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COMPRESSION:Gk characteristic dead loadQk characteristic imposed loadWk characteristic wind loadγf partial safety factor for loadγm partial safety factor for materialsfk characteristic compressive strength of

masonryN ultimate design vertical loadNR ultimate design vertical load resistance of

wall

FLEXURE:fkx par characteristic flexural strength of masonry

with plane of failure parallel to bed jointfkx perp characteristic flexural strength of masonry

with plane of failure perpendicular to bedjoint

α bending moment coefficientµ orthogonal ratioM ultimate design momentMR design moment of resistanceMpar design moment with plane of failure

parallel to bed jointMperp design moment with plane of failure

perpendicular to bed jointMk par design moment of resistance with plane

of failure parallel to bed jointMk perp design moment of resistance with plane

of failure perpendicular to bed joint

5.5 Design of vertically loadedmasonry walls

In common with most modern codes of practicedealing with structural design, BS 5628: Code ofPractice for Use of Masonry is based on the limit

Fig. 5.8 Masonry walls.

The ultimate design load is a function of the actualloads bearing down on the wall. The design loadresistance is related to the design strength of themasonry wall. The following sub-sections discussthe procedures for estimating the:

1. ultimate design load,2. design strength of masonry walls and3. design load resistance of masonry walls.

5.5.1 ULTIMATE DESIGN LOADS, NAs discussed in section 2.3, the loads acting ona structure can principally be divided into threebasic types, namely dead loads, imposed (or live)loads and wind loads. Generally, the ultimate designload is obtained by multiplying the characteristic(dead/imposed/wind) loads (Fk) by the appropriatepartial safety factor for loads (γf)

N = γfFk (5.2)

state philosophy (Chapter 1). This code states thatthe primary aim of design is to ensure an adequatemargin of safety against the ultimate limit statebeing reached. In the case of vertically loaded wallsthis is achieved by ensuring that the ultimatedesign load (N ) does not exceed the design loadresistance of the wall (NR):

N ≤ NR (5.1)

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Design of vertically loaded masonry walls

5.5.1.1 Characteristic loads (clause 17,BS 5628)

The characteristic values of dead loads (Gk) andimposed loads (Qk) are obtained from the following:(i) BS 648: Schedule of Weights for Building Materials;(ii) BS 6399: Design Loadings for Buildings, Part 1:Code of Practice for Dead and Imposed Loads, Part 3:Code of Practice for Imposed Roof Loads. The charac-teristic wind load (Wk) is calculated in accordancewith BS 6399: Part 2: Wind loads.

5.5.1.2 Partial safety factors for loads (γ f)(clause 18, BS 5628)

As discussed in section 2.3.2, the applied loadsmay be greater than anticipated for a number ofreasons. Therefore, it is normal practice to factorup the characteristic loads. Table 5.8 shows thepartial safety factors on loading for various loadcombinations. Thus, with structures subject to onlydead and imposed loads the partial safety factorsfor the ultimate limit state are usually taken to be1.4 and 1.6 respectively. The ultimate design loadfor this load combination is given by

N = 1.4Gk + 1.6Qk (5.3)

In assessing the effect of dead, imposed and windload for the ultimate limit state, the partial safetyfactor is generally taken to be 1.2 for all the loadtypes. Hence

N = 1.2(Gk + Qk + Wk) (5.4)

5.5.2 DESIGN COMPRESSIVE STRENGTHThe design compressive strength of masonry is given by

Design compressive strength =

βγfk

m

(5.5)

wherefk characteristic compressive strength of masonryγm partial safety factor for materials subject to

compressionβ capacity reduction factor

Each of these factors is discussed in the followingsections.

5.5.2.1 Characteristic compressive strengths ofmasonry, fk (clause 19, BS 5628)

The basic characteristic compressive strengths ofnormally bonded masonry constructed under lab-oratory conditions and tested at an age of 28 daysare given in Table 5.9. As will be noted from thetable, the compressive strength is a function ofseveral factors including the type and size ofmasonry unit, percentage of formed voids or frogs,compressive strength of the unit and designation/strength class of the mortar used. The followingdiscusses how the characteristic strength of ma-sonry constructed from (a) bricks and (b) concreteblocks is determined using Table 5.9.

(a) Brickwork. The basic characteristic strengthsof masonry built with standard format bricks andmortar designations (i)–(iv) (Table 5.6) are obtainedfrom Table 5.9(a). Standard format bricks arebricks having no more than 25 per cent of formedvoids or 20 per cent frogs. The values in the tablemay be modified where the following conditions,amongst others, apply:

1. If the horizontal cross-sectional area of theloaded wall (A) is less than 0.2 m2, the basiccompressive strength should be multiplied bythe factor (0.7 + 1.5A) (clause 19.1.2, BS 5628).

2. When brick walls are constructed so that thethickness of the wall or loaded inner leaf of acavity wall is equal to the width of a standardformat brick, the characteristic compressivestrength obtained from Table 5.9(a) may bemultiplied by 1.15 (clause 19.1.3, BS 5628).

(b) Blockwork. Table 5.9(c), (d) and (f ) areused, singly or in combination, to determine thebasic characteristic compressive strength of con-crete blockwork masonry. As in the case of brickmasonry, the values given in these tables may be

Table 5.8 Partial factors of safety on loading, γ f, for various load combinations

Load combinations Ultimate limit state

Dead Imposed Wind

Dead and imposed 1.4Gk or 0.9Gk 1.6Q k

Dead and wind 1.4Gk or 0.9Gk The larger of 1.4Wk or 0.015Gk

Dead, imposed and wind 1.2Gk 1.2Q k The larger of 1.2Wk or 0.015Gk

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Table 5.9 Characteristic compressive strength of masonry, fk (based on Table 2, BS 5628)(a) Constructed with standard format bricks of clay and calcium silicate

Mortar strength Compressive strength of unit (N/mm2)class/designation

5 10 15 20 30 40 50 75 100 125 150

(i) 2.5 4.0 5.3 6.4 8.3 10.0 11.6 15.2 18.3 21.2 23.9(ii) 2.5 3.8 4.8 5.6 7.1 8.4 9.5 12.0 14.2 16.1 17.9(iii) 2.5 3.4 4.3 5.0 6.3 7.4 8.4 10.5 12.3 14.0 15.4(iv) 2.2 2.8 3.6 4.1 5.1 6.1 7.1 9.0 10.5 11.6 12.7

(c) Constructed with aggregate concrete blocks having a height to least horizontal dimensions of 0.6


2.9 3.6 5.2 7.3 10.4 17.5 22.5 30 40 orgreater

(i) 1.4 1.7 2.5 3.4 4.4 6.3 7.5 9.5 11.2(ii) 1.4 1.7 2.5 3.2 4.2 5.5 6.5 7.9 9.3(iii) 1.4 1.7 2.5 3.2 4.1 5.1 6.0 7.2 8.2(iv) 1.4 1.7 2.2 2.8 3.5 4.6 5.3 6.2 7.1

(d) Constructed with aggregate concrete blocks having not more than 25 per cent of formed voids(i.e. solid) and a ratio of height to least horizontal dimensions of between 2 and 4.5


2.9 3.6 5.2 7.3 10.4 17.5 22.5 30 40 orgreater

(i) 2.8 3.5 5.0 6.8 8.8 12.5 15.0 18.7 22.1(ii) 2.8 3.5 5.0 6.4 8.4 11.1 13.0 15.9 18.7(iii) 2.8 3.5 5.0 6.4 8.2 10.1 12.0 14.5 16.8(iv) 2.8 3.5 4.4 5.6 7.0 9.1 10.5 12.5 14.5

(f ) Constructed with aggregate concrete blocks having more than 25 per cent but less than 60 per centof formed voids (i.e. hollow blocks) and a ratio of height to least horizontal dimensions of between2.0 and 4.5

Mortar Compressive strength of unit (N/mm2)designation

2.8 3.5 5.0 7.0 10 15 20 35 orgreater

(i) 2.8 3.5 5.0 5.7 6.1 6.8 7.5 11.4(ii) 2.8 3.5 5.0 5.5 5.7 6.1 6.5 9.4(iii) 2.8 3.5 5.0 5.4 5.5 5.7 5.9 8.5(iv) 2.8 3.5 4.4 4.8 4.9 5.1 5.3 7.3

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5.5.2.2 Partial safety factor for materials (γm)Table 5.10 shows recommended values of thepartial safety factor for materials, γm, for use indetermining the design compressive strength ofmasonry. The table also gives values of γm to beused in assessing the design strength of masonry inflexural. Note that the values of γm used for flexureare generally lower than the corresponding valuesfor compression. This is different from the 1978edition of BS 5628 in which they are identical.

It will be seen from Table 5.10 that γm dependson (i) the quality of the masonry units and (ii) thequality control during construction. Moreover,there are two levels of control in each case. Thequality of masonry units may be Category I orCategory II, which indicates the consistency of thestrength of the units. Category I units are thosewith a declared compressive strength with a prob-ability of failure to reach that strength not exceed-ing 5 per cent whereas Category II units are thosethat do not conform to this condition. It is themanufacturer’s responsibility to declare the categoryof the masonry units supplied. The categories ofmasonry construction control are ‘normal’ and ‘spe-cial’. For design purposes it is usual to assume thatthe category of construction control is ‘normal’unless the provisions outlined in clause 23.2.2.2(BS 5628) relating to the ‘special’ category canclearly be met.

5.5.2.3 Capacity reduction factor (β)Masonry walls which are tall and slender arelikely to be less stable under compressive loadingthan walls which are short and stocky. Similarly,eccentric loading will reduce the compressive loadcapacity of the wall. In both cases, the reduction inload capacity reflects the increased risk of failuredue to instability rather than crushing of thematerials. These two effects are taken into accountby means of the capacity reduction factor, β(Table 5.11). This factor generally reduces the

Fig. 5.9 Principal dimensions of blocks.

modified where the cross-sectional area of loadedwall is less than 0.2 m2 (see above).

Table 5.9(c) applies to masonry constructedusing any type of aggregate concrete block havinga ratio of height to least horizontal dimension of0.6 (Fig. 5.9). Table 5.9(d) applies to masonry con-structed using solid concrete blocks (i.e. blockshaving not more than 25 per cent of formed voids)and a ratio of height to least horizontal dimensionof between 2 and 4.5. Table 5.9(f) applies to ma-sonry constructed using hollow blocks (i.e. blockshaving more than 25 per cent but less than 60 percent of formed voids) and a ratio of height to leasthorizontal dimension of between 2 and 4.5.

When walls are constructed using aggregate con-crete blocks having less than 25 per cent of formedvoids and a ratio of height to least horizontal dimen-sion of between 0.6 and 2.0, the characteristicstrengths should be determined by interpolatingbetween the values in Table 5.9(c) and (d).

When walls are constructed using aggregate con-crete blocks having more than 25 per cent and lessthan 60 per cent of formed voids and a ratio ofheight to least horizontal dimension of between0.6 and 2.0, the characteristic strengths should bedetermined by interpolating between the values inTable 5.9(c) and 5.9(f).


Table 5.10 Partial safety factors for materials, γm (Table 4,BS 5628)

Category of masonry units Category ofconstruction control

Special Normal

Compression, γm Category I 2.5 3.1Category II 2.8 3.5

Flexure, γm Category I and II 2.5 3.0

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lower slenderness ratios and hence higher designstrengths than walls which are partially fixed orunrestrained.

As can be appreciated from the above, theslenderness ratio depends upon the height of themember, the cross-sectional area of loaded walland the restraints at the member ends. BS 5628defines slenderness ratio, SR, as:

SR

effective heighteffective thickness

ef

ef

= =ht

(5.6)

The following discusses how the (i) effective heightand (ii) effective thickness of masonry walls aredetermined.

(i) E (hef). The effective height ofa load-bearing wall depends on the actual heightof the member and the type of restraint at thewall ends. Such restraint that exists is normallyprovided by any supported floors, roofs, etc. andmay be designated ‘simple’ (i.e. pin-ended) or‘enhanced’ (i.e. partially fixed), depending on theactual construction details used. Figures 5.10 and5.11 show typical examples of horizontal restraintswhich provide simple and enhanced resistancerespectively.

According to clause 24.3.2.1 of BS 5628 theeffective height of a wall may be taken as:

1. 0.75 times the clear distance between lateralsupports which provide enhanced resistance tolateral movements; or

2. the clear distance between lateral supports whichprovide simple resistance to lateral movement.

(ii) E (tef). For single leaf wallsthe effective thickness is taken as the actual thick-ness (t) of the wall (Fig. 5.12(a)):

tef = t (single leaf wall)

Table 5.11 Capacity reduction factor, β(Table 7, BS 5628)

Slenderness ratio Eccentricity at top of wall, ex, up to

hef /tef 0.05t 0.1t 0.2t 0.3t

0 1.00 0.88 0.66 0.446 1.00 0.88 0.66 0.448 1.00 0.88 0.66 0.4410 0.97 0.88 0.66 0.4412 0.93 0.87 0.66 0.4414 0.89 0.83 0.66 0.4416 0.83 0.77 0.64 0.4418 0.77 0.70 0.57 0.4420 0.70 0.64 0.51 0.3722 0.62 0.56 0.43 0.3024 0.53 0.47 0.3426 0.45 0.3827 0.40 0.33

Fig. 5.10 Details of horizontal supports providing simple resistance.

design compressive strength of elements such aswalls and columns, in some cases by as much as70 per cent, and is a function of the following fac-tors which are discussed below: (a) slendernessratio and (b) eccentricity of loading.

(a) Slenderness ratio (SR). The slendernessratio indicates the type of failure which may arisewhen a member is subject to compressive loading.Thus, walls which are short and stocky will have alow slenderness ratio and tend to fail by crushing.Walls which are tall and slender will have higherslenderness ratios and will also fail by crushing.However, in this case, failure will arise as a resultof excessive bending of the wall rather than directcrushing of the materials. Similarly, walls whichare rigidly fixed at their ends will tend to have

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Fig. 5.11 Details of horizontal supports providingenhanced resistance.

If the wall is stiffened with piers (Fig. 5.12(b)) inorder to increase its load capacity, the effectivethickness is given by

tef = tK (single leaf wall stiffened with pier)

where K is obtained from Table 5.12.For cavity walls the effective thickness should be

taken as the greater of (i) two-thirds the sum of theactual thickness of the two leaves, i.e. 2/3(t1 + t2))or (ii) the actual thickness of the thicker leaf, i.e.t1 or t2 (Fig. 5.13(a)). Where the cavity wall isstiffened by piers, the effective thickness of the wallis taken as the greatest of (i) 2/3 (t1 + Kt2) or (ii) t1

or (iii) Kt2 (Fig. 5.13b).

(b) Eccentricity of vertical loading. In addi-tion to the slenderness ratio, the capacity reduc-tion factor is also a function of the eccentricity ofloading. When considering the design of walls itis not realistic to assume that the loading will beapplied truly axially but rather that it will occur atsome eccentricity to the centroid of the wall. Thiseccentricity is normally expressed as a fraction ofthe wall thickness.

Clause 27 of BS 5628 recommends that the loadstransmitted to a wall by a single floor or roof actsat one third of the length of the bearing surfacefrom the loaded edge (Fig. 5.14). This is based onthe assumption that the stress distribution underthe bearing surface is triangular in shape as shownin Fig. 5.14.

Where a uniform floor is continuous over a wall,each span of the floor should be taken as beingsupported individually on half the total bearing area(Fig. 5.15).

5.5.3 DESIGN VERTICAL LOAD RESISTANCE OFWALLS (NR)

The foregoing has discussed how to evaluate thedesign compressive strength of a masonry wallbeing equal to the characteristic strength ( fk)multiplied by the capacity reduction factor (β)and divided by the appropriate factor of safety formaterials (γm). The characteristic strengths andfactors of safety for materials are obtained fromTables 5.9 and 5.10 respectively. The effective thick-ness and effective height of the member are usedto determine the slenderness ratio and thence,together with the eccentricity of loading, thecapacity reduction factor via Table 5.11.

The design strength is used to estimate thevertical load resistance of a wall per metre length,NR, which is given by

Table 5.12 Stiffness coefficient for wallsstiffened by piers (Table 5, BS 5628)

Ratio of pier spacing Ratio tp/t of pier thickness(centre to centre) to actual thickness ofto pier width wall to which it

is bonded

1 2 3

6 1.0 1.4 2.010 1.0 1.2 1.420 1.0 1.0 1.0

Note: Liner interpolation between the values given inthe table is permissible, but not extrapolation outside thelimits given.

Fig. 5.13 Effective thickness: (a) cavity wall; (b) cavitywall stiffened with piers (based on Fig. 2 of BS 5268).

Fig. 5.12 Effective thickness: (a) single leaf wall;(b) single leaf wall stiffened with piers (based on Fig. 2 ofBS 5628).


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Fig. 5.14 Eccentricity for wall supporting single floor.

Fig. 5.15 Eccentricity for wall supporting continuous floor.

NR design compression stress area = ×

= × × =

βγ

βγ

ft

tfk

m

k

m

1 (5.7)

As stated earlier, the primary aim of design isto ensure that the ultimate design load does notexceed the design load resistance of the wall. Thusfrom equations 5.1, 5.2 and 5.7:

N ≤ NR

γf(Gk plus Qk, and/or Wk) ≤

βγt fkm

(5.8)

Equation 5.8 provides the basis for the design ofvertically loaded walls. The full design procedureis summarised in Fig. 5.16.

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Fig. 5.16 Design procedure for vertically loaded walls.

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Example 5.1 Design of a load-bearing brick wall (BS 5628)The internal load-bearing brick wall shown in Fig. 5.17 supports an ultimate axial load of 140 kN per metre runincluding self-weight of the wall. The wall is 102.5 mm thick and 4 m long. Assuming the masonry units conform toCategory II and the construction control category is ‘normal’, design the wall.

LOADINGUltimate design load, N = 140 kN m−1 = 140 N mm−1

DESIGN VERTICAL LOAD RESISTANCE OF WALL

Characteristic compressive strengthBasic value = fk

Check modification factor

Small plan area – modification factor does not apply since horizontal cross-sectional area of wall, A = 0.1025 × 4.0= 0.41 m2 > 0.2 m2.

Narrow brick wall – modification factor is 1.15 since wall is one brick thick.

Hence modified characteristic compressive strength is 1.15fk

Safety factor for materials (γγγγγm)Manufacture and construction controls categories are, respectively, ‘II’ and ‘normal’. Hence from Table 5.10, γm forcompression = 3.5

Capacity reduction factor (βββββ)EccentricitySince wall is axially loaded assume eccentricity of loading, ex < 0.05t

Slenderness ratio (SR)Concrete slab provides ‘enhanced’ resistance to wall:

hef = 0.75 × height = 0.75 × 2800 = 2100 mm

tef = actual thickness (single leaf) = 102.5 mm

SR permissibleef

ef

.

. = = = < =ht

2100102 5

10 05 27

Hence, from Table 5.11, β = 0.68.

Fig. 5.17

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Example 5.1 continuedDesign vertical load resistance of wall (NR)

N

tR

m

modified characteristic strength

=

× ×βγ

=

× ×

. ( . ) ..

0 68 1 15 102 53 5

fk

DETERMINATION OF fkFor structural stability

NR ≥ N

0 68 1 15 102 53 5

. ( . ) ..

× ×fk ≥ 140

Hence

fk

.≥

14022 9

= 6.1N mm−2

SELECTION OF BRICK AND MORTAR TYPEFrom Table 5.9(a), any of the following brick/mortar combinations would be appropriate:

Compressive strength Mortar designation fk

of bricks (N mm−2) (N mm−2)

40 (iv) 6.130 (iii) 6.320 (i) 6.4

Example 5.2 Design of a brick wall with ‘small’ plan area (BS 5628)Redesign the wall in Example 5.1 assuming that it is only 1.5 m long.

The calculations for this case are essentially the same as for the 4 m long wall except for the fact the plan areaof the wall, A, is now less than 0.2 m2, being equal to 0.1025 × 1.5 = 0.1538 m2. Hence, the ‘plan area’ modificationfactor is equal to

(0.70 + 1.5A) = 0.7 + 1.5 × 0.1538 = 0.93

The 1.15 factor applicable to narrow brick walls is still appropriate, and therefore the modified characteristiccompressive strength of masonry is given by

0.93 × 1.15f k = 1.07f k

The actual brick type that will be specified on the working drawings will depend not only upon the structuralrequirements but also durability, acoustics, fire resistance, buildability and cost, amongst others. In this particularcase the designer may specify the minimum requirements as HD clay units, compressive strength 30 N mm−2, F0 (i.e.passive environment) and S0 (i.e. no limit on soluble salt content) in mortar designation (iii). Assuming the wall willbe plastered on both sides, the appearance of the bricks is not an issue.


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Example 5.2 continuedFrom the above γm = 3.5 and β = 0.68. Hence, the required characteristic compressive strength of masonry, fk, isgiven by

0 68 1 07 102 53 5

. ( . ) ..

× ×fk ≥ 140 kNm−1

Hence

fk

.≥

14021 3

= 6.6 N mm−2

From Table 5.9a, any of the following brick/mortar combinations would be appropriate:

ASSUMING ‘ENHANCED’ RESISTANCE

Characteristic compressive strength (fk)From Table 5.6, a 1:1:6 mix corresponds to mortar designation (iii). Since the compressive strength of the bricks is30 N mm−2, this implies that fk = 6.3 N mm−2 (Table 5.9a)

Safety factor for materials (γγγγγm)γm = 3.5

Example 5.3 Analysis of brick walls stiffened with piers (BS 5628)A 3.5 m high wall shown in cross-section in Fig. 5.18 is constructed from clay bricks having a compressive strengthof 30 N mm−2 laid in a 1:1:6 mortar. Calculate the ultimate load-bearing capacity of the wall assuming the partialsafety factor for materials is 3.5 and the resistance to lateral loading is (A) ‘enhanced’ and (B) ‘simple’.

Fig. 5.18



50 (iv) 7.140 (iii) 7.430 (ii) 7.1

Hence it can be immediately seen that walls having similar construction details but a plan area of < 0.2 m2 will havea lower load-carrying capacity.

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Example 5.3 continuedCapacity reduction factor (βββββ)EccentricityAssume wall is axially loaded. Hence ex < 0.05t

Slenderness ratio (SR)With ‘enhanced’ resistance

hef = 0.75 × height = 0.75 × 3500 = 2625mm

Pier spacingPier width

.= =4500440

10 2

Pier thicknessThickness of wall

.= =440215

2 0

Hence from Table 5.12, K = 1.2. The effective thickness of the wall, tef, is equal to

tef = tK = 215 × 1.2 = 258 mm

SR ef

ef

= =ht

2625258

= 10.2 < permissible = 27

Hence, from Table 5.11, β = 0.96.

Design vertical load resistance of wall (NR)

NR =

βγf tkm

. .

.=

× ×0 96 6 3 2153 5

= 371 N mm−1 run of wall

= 371 kN m−1 run of wall

Hence the ultimate load capacity of wall, assuming enhanced resistance, is 371 kN m−1 run of wall.

ASSUMING ‘SIMPLE’ RESISTANCEThe calculation for this case is essentially the same as for ‘enhanced’ resistance except

hef = actual height = 3500 mm

tef = 258 mm (as above)

Hence

SR =

ht

ef

ef

.= =3500258

13 6

Assuming ex < 0.05t (as above), this implies that β = 0.9 (Table 5.11). The design vertical load resistance of the wall,NR, is given by:

NR =

βγf tkm

. .

.=

× ×0 9 6 3 2153 5

= 348 N mm−1 run of wall

= 348 kN m−1 run of wall

Hence it can be immediately seen that, all other factors being equal, walls having simple resistance have a lowerresistance to failure.


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Example 5.4 Design of single leaf brick and block walls (BS 5628)Design the single leaf load-bearing wall shown in Fig. 5.19 using mortar designation (iii) and either (a) standardformat bricks or (b) solid concrete blocks of length 390 mm, height 190 mm and thickness 100 mm. Assume themasonry units conform to Category I and the construction control category is ‘normal’.

DESIGN FOR STANDARD FORMAT BRICKS

Loading

Characteristic dead load gkAssuming that the bricks are made from clay, the density of the brickwork can be assumed to be 55 kg m−2 per 25 mmthickness (Table 2.1) or (55 × (102.5/25) 9.81 × 10−3 =) 2.2 kN m−2. Therefore self-weight of wall is

SW = 2.2 × 2.95 × 1 = 6.49 kN/metre run of wall.

Dead load due to roof =

( ) .5 5 4 82

+ × = 24 kN m−1 run

gk = SW + roof load

= 6.49 + 24 = 30.49 kN m−1 run of wall

Characteristic imposed load, qk

qk = roof load

=

( ) .5 5 1 52

+ × = 7.5 kN m−1 run of wall

Ultimate design load, N

N = 1.4gk + 1.6qk

= 1.4 × 30.49 + 1.6 × 7.5 = 54.7 kN m−1 run of wall

Design vertical load resistance of wall

Characteristic compressive strength

Basic value = fk

Fig. 5.19

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Example 5.4 continuedCheck modification factor

Small plan area – modification factor will not apply provided loaded plan area A > 0.2 m2, i.e provided that the walllength exceeds 2 m ≈ 0.2 m2/0.1025 m

Narrow brick wall – since wall is one brick thick, modification factor = 1.15

Hence, modified characteristic compressive strength = 1.15fk

Safety factor for materials (γγγγγm)Category of masonry unit is ‘I’ and the category of construction control is ‘normal’. Hence from Table 5.10 γm = 3.1.

Capacity reduction factor (βββββ)

EccentricityAssuming the wall is symmetrically loaded, eccentricity of loading, ex < 0.05t


hef = 0.75 × height = 0.75 × 2950 = 2212.5 mm

tef = actual thickness (single leaf) = 102.5 mm

SR =

ht

ef

ef

.

. =

2212 5102 5

= 21.6

Hence from Table 5.11, β = 0.63.


NR =

βγ

× ×modified characteristic strength

m

t

=

0 63 1 15 102 53 1

. ( . ) ..

× ×fk

Determination of fkFor structural stability

NR ≥ N

0 63 1 15 102 53 1

. ( . ) ..

× ×fk ≥ 54.7

Hence

fk N mm

.

. . ≥ = −54 7

23 952 3 2

Selection of brick and mortar typeFrom Table 5.9(a), any of the following brick/mortar combinations would be appropriate:



10 (iv) 2.85 (iii) 2.5


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Example 5.4 continuedDESIGN FOR SOLID CONCRETE BLOCKS

LoadingAssuming that the density of the blockwork is the same as that for brickwork, i.e. 55 kg m−2 per 25 mm thickness(Table 2.1), self-weight of wall is 6.49 kN m−1 run of wall.

Dead load due to roof = 24 kN m−1 run of wall

Imposed load from roof = 7.5 kN m−1 run of wall

Ultimate design load, N = 1.4(6.49 + 24) + 1.6 × 7.5

= 54.7 kN per m run of wall

Design vertical load resistance of wall

Characteristic compressive strengthCharacteristic strength is fk assuming that plan area of wall is greater than 0.2 m2. Note that the narrow wallmodification factor only applies to brick walls.

Safety factor for materials

γm = 3.1

Capacity reduction factorEccentricity

ex < 0.05t

Slenderness ratio (SR)Concrete slab provides ‘enhanced’ resistance to wall

hef = 0.75 × height = 0.75 × 2950 = 2212.5 mm

tef = actual thickness (single leaf) = 100 mm

SR =

ht

ef

ef

.

.= =2212 5

10022 1

Hence from Table 5.11, β = 0.61.


NR =

βγ

× ×modified characteristic strength

m

t

=

0 61 1003 1

. ( ) .

× ×fk

Determination of fkFor structural stability

NR ≥ N

0 61 1003 1

54 7.

. .

× ×≥

fk

Hence

fk N mm

.

. . ≥ = −54 7

19 682 8 2

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Example 5.4 continuedSelection of block and mortar type

height of block

least horizontal dimension of block (i.e. thickness) .= =

190

1001 9

Ratio of height to least 0.6 1.9 2.0horizontal dimensionCompressive strength of 1.7 3.3 3.5masonry, fk (Nmm−2) (from Table 5.9c) (from Table 5.9d )

Interpolating between Tables 5.9(c) and (d) a solid block of compressive strength 3.6 N mm−2 used with mortardesignation (iv) would produce masonry of approximate compressive strength 3.3 N mm−2 which would be appro-priate here.

Example 5.5 Design of a cavity wall (BS 5628)A cavity wall of length 6 m supports the loads shown in Fig. 5.20. The inner load-bearing leaf is built using concreteblocks of length 440 mm, height 215 mm, thickness 100 mm and faced with plaster, and the outer leaf from standardformat clay bricks. Design the wall assuming the masonry units are category I and the construction control categoryis normal. The self-weight of the blocks and plaster can be taken to be 2.4 kN m−2.

Clause 25.2.1 of BS 5628 on cavity walls states that ‘where the load is carried by one leaf only, the load-bearingcapacity of the wall should be based on the horizontal cross-sectional area of that leaf alone, although the stiffeningeffect of the other leaf can be taken into account when calculating the slenderness ratio’. Thus, it should be notedthat the following calculations relate to the design of the inner load-bearing leaf.

LOADING

Characteristic dead load gk

gk = roof load + self weight of wall

=

( . ) 6 5 1 62

× × + (3.5 × 1) × 2.4

= 19.5 + 8.4 = 27.9 kN per m run of wall

Fig. 5.20


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Example 5.5 continuedCharacteristic imposed load, qk

qk = roof load

=

( . ) .6 5 11 52×

= 4.9 kN per m run of wall

Ultimate design load, NN = 1.4gk + 1.6qk

= 1.4 × 27.9 + 1.6 × 4.9 = 46.9 kN m−1 run of wall

DESIGN VERTICAL LOAD RESISTANCE OF WALL

Characteristic compressive strengthBasic value = fk

Check modification factorsSmall plan area – modification factor does not apply since loaded plan area, A > 0.2 m2 where A = 6 × 0.1 = 0.6 m2

Narrow brick wall – modification factor does not apply since inner leaf wall is blockwork

Safety factor for materialsMasonry units are category I and category of masonry construction control is normal. Hence, from Table 5.10,γm = 3.1.

Capacity reduction factorEccentricityThe load from the concrete roof will be applied eccentrically as shown in the figure. The eccentricity is given by:

ex =

t t tt

2 3 60 167 .− = =

SR =

ht

ef

ef

=2625135



hef = 0.75 × height = 0.75 × 3500 = 2625 mm

tef = 135 mm

since tef is the greater of

2/3(t1 + t2)[= 2/3(102.5 + 100) = 135 mm] and t1 [= 102.5 mm] or t2 [= 100 mm]

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Hence by interpolating between the values in Table 5.11, β = 0.57.


NR =

βγf t fk

m

k .

.=

0 57 1003 1

DETERMINATION OF fk

For structural stability

NR ≥ N

0 57 1003 1

46 9.

. .

fk ≥

Hence

fk

..

≥46 9

18 39 = 2.6 N mm−2

SELECTION OF BLOCK, BRICK AND MORTAR TYPE

height

thicknessof block .= =

215

1002 15

From Table 5.9(d), a solid concrete block with a compressive strength of 2.9 N mm−2 used with mortar type (iv) wouldbe appropriate. The bricks and mortar for the outer leaf would be selected on the basis of appearance and durability(Table 5.7).

5.6 Design of laterally loadedwall panels

A non-load-bearing wall which is supported on anumber of sides is usually referred to as a panelwall. Panel walls are very common in the UK andare mainly used to clad framed buildings. Thesewalls are primarily designed to resist lateral loadingfrom the wind.

The purpose of this section is to describe thedesign of laterally loaded panel walls. This requiresan understanding of the following factors whichare discussed below:

1. characteristic flexural strength2. orthogonal ratio3. support conditions4. limiting dimensions5. basis of design

5.6.1 CHARACTERISTIC FLEXURAL STRENGTHOF MASONRY (fkx)

The majority of panel walls tend to behave as atwo-way bending plate when subject to lateralloading (Fig. 5.21). However, it is more convenientto understand the behaviour of such walls if thebending processes in the horizontal and verticaldirections are studied separately.

Design of laterally loaded wall panels

Slenderness Eccentricity at top of wall, ex

ratiohef/tef 0.1t 0.167t 0.2t

18 0.70 0.5719.4 0.658 0.570 0.52820 0.64 0.51

SR =

ht

ef

ef

=2625135


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264

Fig. 5.21 Panel wall subject to two-way bending.

Fig. 5.22 Failure modes of walls subject to lateral loading: (a) failure perpendicular to bed joint; (b) failure parallel tobed joint.

Bending tests carried out on panel walls showthat the flexural strength of masonry is significantlygreater when the plane of failure occurs perpen-dicular to the bed joint rather than when failureoccurs parallel to the bed joint (Fig. 5.22). This isbecause, in the former case, failure is a complexmix of shear, etc. and end bearing, while in thelatter case cracks need only form at the mortar/masonry unit interface.

Table 5.13 shows the characteristic flexuralstrengths of masonry constructed using a range ofbrick/block types and mortar designations for thetwo failure modes. Note that for masonry builtwith clay bricks, the flexural strengths are prim-arily related to the water-absorption properties ofthe brick, but for masonry built with concrete blocksthe flexural strengths are related to the compressivestrengths and thicknesses of the blocks.

5.6.2 ORTHOGONAL RATIO, µThe orthogonal ratio is the ratio of the characteristicflexural strength of masonry when failure occursparallel to the bed joints ( fkx par) to that when fail-ure occurs perpendicular to the bed joint ( fkx perp):

µ =

ff

kx par

kx perp

(5.9)

As will be discussed later, the orthogonal ratio isused to calculate the size of the bending momentin panel walls.

For masonry constructed using clay, calciumsilicate or concrete bricks a value of 0.3 for µ cannormally be assumed in design. No such uniquevalue exists for masonry walls built with concreteblocks and must, therefore, be determined for indi-vidual block types.

5.6.3 SUPPORT CONDITIONSIn order to assess the lateral resistance of masonrypanels it is necessary to take into account the sup-port conditions at the edges. Three edge conditionsare possible:

(a) a free edge(b) a simply supported edge(c) a restrained edge.

A free edge is one that is unsupported (Fig. 5.23).A restrained edge is one that, when the panel isloaded, will fail in flexure before rotating. All othersupported edges are assumed to be simply sup-ported. This is despite the fact that some degree offixity may actually exist in practice.

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Table 5.13 Characteristic flexural strength of masonry, fkx in N mm−2 (Table 3, BS 5628)

Plane of failure Plane of failureparallel to bed joints perpendicular to bed joints

Mortar strength M12 M6 and M4 M2 M12 M6 and M4 M2class/designation (i) (ii) and (iii) (iv) (i) (ii) and (iii) (iv)

Clay bricks having a water absorptionless than 7% 0.7 0.5 0.4 2.0 1.5 1.2between 7% and 12% 0.5 0.4 0.35 1.5 1.1 1.0over 12% 0.4 0.3 0.25 1.1 0.9 0.8

Calcium silicate bricks 0.3 0.2 0.9 0.6Concrete bricks 0.3 0.2 0.9 0.6

Concrete blocks (solids or hollow) of compressivestrength in Nmm−2:

2.9 used in walls of 0.40 0.43.6 thicknessa 0.25 0.2 0.45 0.47.3 up to 100 mm 0.60 0.5

2.9 used in walls of 0.40 0.23.6 thicknessa 0.15 0.1 0.45 0.27.3 250 mm of greater 0.60 0.3

10.4 used in walls of any 0.75 0.617.5 thicknessa 0.25 0.2 0.90b 0.7b

and over

aThe thickness should be taken to be the thickness of the wall, for a single-leaf wall, or the thickness of the leaf, for a cavity wall.bWhen used with flexural strength in parallel direction, assume the orthogonal ratio µ = 0.3.

56756798

567567567

Figures 5.24 and 5.25 show typical details pro-viding simple and restrained support conditionsrespectively.

5.6.4 LIMITING DIMENSIONSClause 32.3 of BS 5628 lays down various limitson the dimensions of laterally loaded panel wallsdepending upon the support conditions. These arechiefly in respect of (i) the area of the panel and(ii) the height and length of the panel. Specifically,the code mentions the following limits:Fig. 5.23 Detail providing free edge.


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266

Fig. 5.24 Details providing simple support conditions: (a) and (b) vertical supports; (c)–(e) horizontal support(based on Figs 6 and 7, BS 5628).

Fig. 5.25 Details providing restrained support conditions: (a) and (b) vertical support; (c) horizontal support at head ofwall (based on Figs 6 and 7, BS 5628).

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267


Fig. 5.26 Panel sizes: (a) panels supported on three sides; (b) panels supported on four edges; (c) panel simply supportedtop and bottom.

1. Panel supported on three edges(a) two or more sides continuous:

height × length equal to 1500t 2ef or less;

(b) all other cases:height × length equal to 1350t 2

ef or less.2. Panel supported on four edges

(a) three or more sides continuous:height × length equal to 2250t 2

ef or less;(b) all other cases:

height × length equal to 2025t 2ef or less.

3. Panel simply supported at top and bottomHeight equal to 40tef or less.

Figure 5.26 illustrates the above limits on panelsizes. The height and length restrictions referredto in (ii) above only apply to panels which aresupported on three or four edges, i.e. (1) and (2). Insuch cases the code states that no dimension shouldexceed 50 times the effective thickness of the wall(tef).

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268

5.6.5 BASIS OF DESIGN (CLAUSE 32.4,BS 5628)

The preceding sections have summarised much ofthe background material needed to design laterallyloaded panel walls. This section now considers thedesign procedure in detail.

The principal aim of design is to ensure that theultimate design moment (M) does not exceedthe design moment of resistance of the panel(Md):

M ≤ Md (5.10)

Since failure may take place about either axis(Fig. 5.21), for a given panel, there will be twodesign moments and two corresponding momentsof resistance. The ultimate design moment per unitheight of a panel when the plane of failure is per-pendicular to the bed joint, Mperp (Fig. 5.22), isgiven by:

Mperp = αWkγ fL2 (5.11)

The ultimate design moment per unit height of apanel when the plane of bending is parallel to thebed joint, Mpar, is given by

Mpar = µαWkγ fL2 (5.12)

whereµ orthogonal ratioα bending moment coefficient taken from

Table 5.14γ f partial safety factor for loads

(Table 5.8)L length of the panel between supportsWk characteristic wind load per unit area

The corresponding design moments of resist-ance when the plane of bending is perpendicular,Mk perp, or parallel, Mk par, to the bed joint is givenby equations 5.13 and 5.14 respectively:

Mk perp =

f Zkx perp

mγ(5.13)

Mk par =

f Zkx par

mγ(5.14)

wherefkx perp characteristic flexural strength

perpendicular to the plane of bendingfkx par characteristic flexural strength parallel to

the plane of bending (Table 5.13)Z section modulusγm partial safety factor for materials

(Table 5.10)

Equations 5.10–5.14 form the basis for thedesign of laterally loaded panel walls. It should benoted, however, since by definition µ = fkx par /fkx perp,Mk par = µMk perp (by dividing equation 5.14 by equa-tion 5.13) and Mpar = µMperp (by dividing equation5.11 by equation 5.12) that either equations 5.11and 5.13 or equations 5.12 and 5.14 can be usedin design. These equations ignore the presence ofvertical loading acting on panel walls. The verticalloading arises from the self-weight of the wall anddead load from upper levels in multi-storey build-ings. The presence of vertical loading will enhancethe flexural strength parallel to bed joints as thetendency for flexural tension failure is reduced. Thefull design procedure is summarised in Fig. 5.27and illustrated by means of the following examples.

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269


Table 5.14 Bending moment coefficients in laterally loaded wall panels (based on Table 8, BS 5628)

Values of αµ h/L

0.30 0.50 0.75 1.00 1.25 1.50 1.75

1.00 0.031 0.045 0.059 0.071 0.079 0.085 0.0900.90 0.032 0.047 0.061 0.073 0.081 0.087 0.0920.80 0.034 0.049 0.064 0.075 0.083 0.089 0.0930.70 0.035 0.051 0.066 0.077 0.085 0.091 0.095

A 0.60 0.038 0.053 0.069 0.080 0.088 0.093 0.0970.50 0.040 0.056 0.073 0.083 0.090 0.095 0.0990.40 0.043 0.061 0.077 0.087 0.093 0.098 0.1010.35 0.045 0.064 0.080 0.089 0.095 0.100 0.1030.30 0.048 0.067 0.082 0.091 0.097 0.101 0.1041.00 0.020 0.028 0.037 0.042 0.045 0.048 0.0500.90 0.021 0.029 0.038 0.043 0.046 0.048 0.0500.80 0.022 0.031 0.039 0.043 0.047 0.049 0.0510.70 0.023 0.032 0.040 0.044 0.048 0.050 0.0510.60 0.024 0.034 0.041 0.046 0.049 0.051 0.052

C 0.50 0.025 0.035 0.043 0.047 0.050 0.052 0.0530.40 0.027 0.038 0.044 0.048 0.051 0.053 0.0540.35 0.029 0.039 0.045 0.049 0.052 0.053 0.0540.30 0.030 0.040 0.046 0.050 0.052 0.054 0.0541.00 0.008 0.018 0.030 0.042 0.051 0.059 0.0660.90 0.009 0.019 0.032 0.044 0.054 0.062 0.0680.80 0.010 0.021 0.035 0.046 0.056 0.064 0.0710.70 0.011 0.023 0.037 0.049 0.059 0.067 0.0730.60 0.012 0.025 0.040 0.053 0.062 0.070 0.076E0.50 0.014 0.028 0.044 0.057 0.066 0.074 0.0800.40 0.017 0.032 0.049 0.062 0.071 0.078 0.0840.35 0.018 0.035 0.052 0.064 0.074 0.081 0.0860.30 0.020 0.038 0.055 0.068 0.077 0.083 0.089

Key to support conditions

denotes free edgesimply supported edgean edge over which fullcontinuity exists

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270

Figure 5.27 Design procedure for laterally loaded panel walls.

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271

Example 5.6 Analysis of a one-way spanning wall panel (BS 5628)Estimate the characteristic wind pressure that the cladding panel shown below can resist assuming that it isconstructed using bricks having a water absorption of < 7% and mortar designation (iii). Assume γf = 1.2 and γm = 3.0.

ULTIMATE DESIGN MOMENT (M )Since the vertical edges are unsupported, the panel must span vertically and, therefore, equations 5.11 and 5.12cannot be used to determine the design moment here. The critical plane of bending will be parallel to the bed jointand the ultimate design moment at mid-height of the panel, M, is given by (clause 32.4.2 of BS 5628)

M =

Ultimate load height ×8

Ultimate load on the panel = wind pressure × area

= (γfWk)(height × length of panel)

= 1.2Wk3000 × 1000 = 3.6Wk106 N m−1 length of wall

Hence

M =

3 6 10 30008

6. WK × × = 1.35 × 109 Wk N mm m−1 length of wall

MOMENT OF RESISTANCE (Md)

Section modulus (Z)

Z

bd

. . = =

×= ×

2 3 26 3

610 102 5

61 75 10 mm m−1 length of wall

Moment of resistanceThe design moment of resistance, Md, is equal to the moment of resistance when the plane of bending is parallel tothe bed joint, Mk par. Hence

Md = Mk par =

f Zkx par

mγ

. . .

=× ×0 5 1 75 10

3 0

6

= 0.292 × 106 N mm m−1

where fkx par = 0.5 N mm−2 from Table 5.13, since water absorption of bricks < 7% and the mortar is designation (iii).

DETERMINATION OF CHARACTERISTIC WIND PRESSURE (Wk)For structural stability:

M ≤ Md

1.35 × 109Wk ≤ 0.292 × 106

Wk ≤ 0.216 × 10−3 N mm−2

Hence the characteristic wind pressure that the panel can resist is 0.216 × 10−3 N mm−2 or 0.216 kN m−2.


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272

Example 5.7 Analysis of a two-way spanning panel wall (BS 5628)The panel wall shown in Fig. 5.28 is constructed using clay bricks having a water absorption of greater than 12 percent and mortar designation (ii). If the masonry units are category II and the construction control category is normal,calculate the characteristic wind pressure, Wk, the wall can withstand. Assume that the wall is simply supported onall four edges.

Fig. 5.28

ULTIMATE DESIGN MOMENT (M)Orthogonal ratio (µµµµµ)

µ

.

. .= = =

ff

kx par

kx perp

0 30 9

0 33 (Table 5.13)

Bending moment coefficient (ααααα)

hL

=34

= 0.75

Hence, from Table 5.14(E), α = 0.053

Ultimate design momentM = Mperp = αγfWkL2 = 0.053 × 1.2Wk × 42 kN m m−1 run = 1.0176 × 106Wk N mm m−1 run

MOMENT OF RESISTANCE (Md)Safety factor for materials (γγγγγm)

γm = 3.0 (Table 5.10)

Section modulus (Z)

Z =

bd 2 3 2

610 102 5

6

.=

× = 1.75 × 106 mm3 m−1 length of wall

Moment of resistance (Md)

Md = Mk perp =

f Zkx perp

mγ

. . .

=× ×0 9 1 75 10

3 0

6

= 0.525 × 106 N mm m−1 run

DETERMINATION OF CHARACTERISTIC WIND PRESSURE (Wk)For structural stability

M ≤ Md

1.0176 × 106Wk ≤ 0.525 × 106

⇒ Wk ≤ 0.516 kN m−2

Hence the panel is able to resist a characteristic wind pressure of 0.516 kN m−2.

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273

Example 5.8 Design of a two-way spanning single-leaf panel wall(BS 5628)

A 102.5 mm brick wall is to be designed to withstand a characteristic wind pressure, Wk, of 0.65 kN m−2.Assuming the edge support conditions are as shown in Fig. 5.29 and the masonry units are category II and the

construction control category is ‘normal’, determine suitable brick/mortar combination(s) for the wall.


Fig. 5.29


assume µ = 0.35


hL

=24754500

= 0.55

Hence, from Table 5.14(C), α = 0.04

Ultimate design momentM = Mperp = αγfWkL2 = 0.040 × 1.2 × 0.65 × 4.52 kNm m−1 run

= 0.632 kNm/m run = 0.632 × 106 Nmm m−1 run

MOMENT OF RESISTANCE (Md)Safety factor for materials (γγγγγm)

γm = 3.0 (Table 5.10)

Section modulus (Z)

Z

bd

. . = =

×= ×

2 3 26 3

610 102 5



M M

f Zfd k perp

kx perp

mkx perp

.

.= = =

×γ

1 75 10

3 0

6

= 0.583 × 106 fkx perp N mm m−1 run

DETERMINATION OF fkx perp

For structural stability:

M ≤ Md

0.632 × 106 ≤ 0.583 × 106fkx perp

⇒ fkx perp ≥ 1.08 N mm−2

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274

SELECTION OF BRICK AND MORTAR TYPEFrom Table 5.13, any of the following brick/mortar combination would be appropriate:


Clay brick having a Mortar fkx perp

water absorption: designation (N mm−2)

< 7% (iv) 1.27%–12% (iii) 1.1> 12% (i) 1.1

The design wind pressure, Wk, depends upon the available capacities of both leaves of the wall. Therefore, it is normalpractice to work out the capacities of the outer and inner leaf separately, and to then sum them in order to determineWk.

OUTER LEAFUltimate design moment (M)Orthogonal ratio (µµµµµ)

µ =

ff

kx par

kx perp

..

=0 51 5

= 0.33 (Table 5.13)

Note that the actual value of µ for the above brick/mortar combinations is 0.33 and not 0.35 as assumed. However,this difference is insignificant and will not affect the choice of the brick/mortar combinations shown in the abovetable.

Example 5.9 Analysis of a two-way spanning cavity panel wall(BS 5628)

Determine the characteristic wind pressure, Wk, which can be resisted by the cavity wall shown in Fig. 5.30 if theconstruction details are as follows:

1. Outer leaf: clay brick having a water absorption < 7% laid in a 1:1:6 mortar (i.e. designation (iii)).2. Inner leaf : solid concrete blocks of compressive strength 3.5 N/mm2 and length 390 mm, height 190 mm and

thickness 100 mm also laid in a 1:1:6 mortar.

Assume that the top edge of the wall is unsupported but that the base and vertical edges are simply supported. It canalso be assumed that the masonry units are category II and the construction control category is ‘normal’.

Fig. 5.30

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275

Example 5.9 continuedBending moment coefficient (ααααα)

hL

..

.= =2 55 0

0 5

Hence from Table 5.14(A), α = 0.064

Ultimate design moment

M = Mperp = αγfWkL2 = 0.064 × 1.2(Wk)outer × 52 kN m m−1 run

= 1.92 × 106(Wk)outer Nmm m−1 run

Moment of resistance (Md)Safety factor for materials (γγγγγm)

γm = 3.0 (Table 5.10)

Section modulus (Z )

Z

bd

.= =

×2 3 2

610 102 5

6 = 1.75 × 106 mm3 m−1 length of wall


Md = M k perp =

f Zkx perp

mγ

. .

.=

× ×1 5 1 75 10

3 0

6

= 0.875 × 106 Nmm m−1

Maximum permissible wind pressure on outer leaf (Wk)outerFor structural stability:

M ≤ Md

1.92(Wk)outer × 106 ≤ 0.875 × 106

(Wk)outer ≤ 0.45 Nmm−2 = 0.45 kN m−2

INNER LEAF

Ultimate design moment (M)

Orthogonal ratio (µµµµµ)

µ

.

. .= = =

ff

kx par

kx perp

0 250 45

0 55 (Table 5.13)


hL

..

.= =2 55 0

0 5

Hence from Table 5.14(A), α = 0.055

Ultimate design moment

M = Mperp = αγfWkL2 = 0.055 × 1.2(Wk)inner × 52 kNm m−1 run

= 1.65 × 106(Wk)inner Nmm m−1 run


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276

Example 5.9 continuedMoment of resistance (Md)

Safety factor for materials (γγγγγm)

γm = 3.0 (Table 5.10)

Section modulus (Z )

Z

bd

= =

×2 3 2

610 100


Moment of resistance (Md )

Md = M k perp =

f Zkx perp

mγ

. .

.=

× ×0 45 1 67 10

3 0

6

= 0.25 × 106 Nmm m−1

Maximum permissible wind pressure on inner leaf (Wk)innerFor structural stability:

M ≤ Md

1.65(Wk)inner × 106 ≤ 0.25 × 106 Nmm m−1 run

(Wk)inner ≤ 0.15 Nmm−2 = 0.15 kN m−2

Hence, characteristic wind pressure that the panel can withstand, Wk, is given by:

Wk = (Wk)inner + (Wk)outer

= 0.15 + 0.45

= 0.6 kN m−2

5.7 SummaryThis chapter has explained the basic properties ofthe materials used in unreinforced masonry andthe procedures involved in the design of (a) singleleaf and cavity walls, with and without stiffeningpiers, subject to vertical loading and (b) panel wallsresisting lateral loading. The design proceduresoutlined are in accordance with BS 5628: Part 1:Structural use of Unreinforced Masonry which is basedon limit state philosophy. In the case of vertically

loaded walls it was found that the design processsimply involves ensuring that the ultimate designload does not exceed the design strength. Panelwalls, on the other hand, are normally subject totwo-way bending and the plane of failure mayoccur parallel or perpendicular to bed joints. There-fore, the designer must consider the design mo-ments about both axes and check that they do notexceed the corresponding moment of resistance ofthe panel.

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277

Fig. Q2

Summary

Questions

1. (a) Discuss why there was a decline andwhat factors have led to a revival inthe use of masonry in constructionover recent years.

(b) The 4.0 m high wall show below isconstructed from clay bricks having acompressive strength of 20 Nmm−2

laid in a 1:1:6 mortar. Calculate thedesign load resistance of the wallassuming the partial safety factor formaterials is 3.5 and the resistance tolateral loading is simple.

external wall consists of 102.5 mmthick brick outer leaf with 390 × 190× 100 concrete block. Assume that themanufacturing and constructioncontrols are special and normalrespectively and that the self-weightof the brick and blockwork is2.2 kN m−2.

3. (a) Explain the difference between simpleand enhanced resistance and sketchtypical construction details showingexamples of both types of restraint.

(b) The brick cavity wall shown belowsupports an ultimate axial load of200 kNm−1. Assuming that the load isequally shared by both leaves and themasonry units are category II and thecategory of construction control isnormal, design the wall.

2. (a) Discuss the factors to be consideredin the selection of brick, block andmortar types for particularapplications.

(b) Design the external cavity wall andinternal single leaf load-bearing wallfor the single-storey building shownbelow. Assume that the internal wallis made from solid concrete blocks oflength 390 mm, height 190 mm andthickness 100 mm and that the

102.5 50102.5

326.5

3 m

326.5

102.5

Fig. Q1

Roof dead load (including self weight) = 5.5 kNm−2

Roof imposed load = 1.5 kNm−2

4. (a) Explain with the use of sketches thelimiting dimensions of laterally loadedpanel walls.

Fig. Q3

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278

(b) A 102.5 mm brick wall is to bedesigned to withstand a wind pressureof 550 N m−2. For the supportconditions shown in (i) and (ii) below,determine suitable brick/mortarcombinations for the two cases.Assume the masonry units arecategory II and the category ofconstruction control is normal.

(b) Determine the characteristic windpressure that can be resisted by thecavity wall shown below if theconstruction details are as follows:

1. outer leaf: clay bricks havinga water absorption7%–12% laid in mortardesignation (ii);

2. inner leaf: solid concrete blocksof compressive strength3.6 Nmm−2 and length390 mm, height190 mm and thickness100 mm also laid inmortar designation (ii).

Assume that the masonry units arecategory I, the category of constructioncontrol is normal and the vertical edgesare unsupported.

Fig. Q4

5. (a) In all the examples on the design oflaterally loaded panel wall discussed inthis chapter the self-weight of the wallhas been ignored. What effect do youthink that including the self-weightwould have on the design compressivestrength of masonry? Fig. Q5

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279

This chapter is concerned with the design of timberelements to British Standard 5268: Part 2, which isbased on the permissible stress philosophy. The chapterdescribes how timber is specified for structural purposesand discusses some of the basic concepts involved suchas stress grading, grade stresses and strength classes.The primary aim of this chapter is to give guidance onthe design of flexural members, e.g. beams and joists,compression members, e.g. posts and columns and loadsharing systems, e.g. stud walling.

6.1 IntroductionWood is a very versatile raw material and is stillwidely used in construction, especially in countriessuch as Canada, Sweden, Finland, Norway andPoland, where there is an abundance of good-quality timber. Timber can be used in a rangeof structural applications including marine works:construction of wharves, piers, cofferdams; heavycivil works: bridges, piles, shoring, pylons; do-mestic housing: roofs, floors, partitions; shutteringfor precast and in situ concrete; falsework for brickor stone construction.

Of all the construction materials which havebeen discussed in this book, only timber is naturallyoccurring. This makes it a very difficult material tocharacterise and partly accounts for the wide vari-ation in the strength of timber, not only betweendifferent species but also between timber of thesame species and even from the same log. Quitenaturally, this led to uneconomical use of timberwhich was costly for individuals and the nation asa whole. However, this problem has now beenlargely overcome by specifying stress graded tim-ber (section 6.2).

There is an enormous variety of timber species.They are divided into softwoods and hardwoods, abotanical distinction, not on the basis of mechani-cal strength. Softwoods are derived from trees with

needle-shaped leaves and are usually evergreen,e.g. fir, larch, spruce, hemlock, pine. Hardwoodsare derived from trees with broad leaves and areusually deciduous, e.g. ash, elm, oak, teak, iroko,ekki, greeheart. Obviously the suitability of a par-ticular timber type for any given purpose will de-pend upon various factors such as performance,cost, appearance and availability. This makes speci-fication very difficult. The task of the structuralengineer has been simplified, however, by group-ing timber species into sixteen strength classes forwhich typical design parameters, e.g. grade stressesand moduli of elasticity, have been produced (sec-tion 6.3). Most standard design in the UK is withsoftwoods.

Design of timber elements is normally carriedout in accordance with BS 5268: Structural Use ofTimber. This is divided into the following parts:

Part 2: Code of Practice for Permissible Stress Design,Materials and Workmanship.

Part 3: Code of Practice for Trussed Rafter Roofs.Part 4: Fire Resistance of Timber Structures.Part 5: Code of Practice for the Preservative Treat-

ment of Structural Timber.Part 6: Code of Practice for Timber Frame Walls.Part 7: Recommendations for the Calculation Basis for

Span Tables.

The design principles which will be outlined inthis chapter are based on the contents of Part 2 ofthe code. It should therefore be assumed that allfuture references to BS 5268 refer exclusively toPart 2. As pointed out in Chapter 1 of this book,BS 5268 is based on permissible stress design ratherthan limit state design. This means in practice thata partial safety factor is applied only to materialproperties, i.e. the permissible stresses (section 6.4)and not the loading.

Specifically, this chapter gives guidance on thedesign of timber beams, joists, columns and studwalling. The design of timber formwork is not

Chapter 6

Design in timber toBS 5268

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Design in timber to BS 5268

280

covered here as it was considered to be rather toospecialised a topic and, therefore, inappropriatefor a book of this nature. However, before discuss-ing the design process in detail, the followingsections will expand on the more general aspectsmentioned above, namely:

a) stress gradingb) grade stress and strength classc) permissible stress.

6.2 Stress gradingThe strength of timber is a function of severalparameters including the moisture content, density,size of specimen and the presence of variousstrength-reducing characteristics such as knots,slope of grain, fissures and wane. Prior to the in-troduction of BS 5268 the strength of timber wasdetermined by carrying out short-term loading testson small timber specimens free from all defects.The data were used to estimate the minimumstrength which was taken as the value below whichnot more than 1% of the test results fell. Thesestrengths were multiplied by a reduction factor togive basic stresses. The reduction factor made anallowance for the reduction in strength due toduration of loading, size of specimen and othereffects normally associated with a safety factor suchas accidental overload, simplifying assumptionsmade during design and design inaccuracies,together with poor workmanship. Basic stress wasdefined as the stress which could safely be perman-ently sustained by timber free from any strength-reducing characteristics. Basic stress, however, wasnot directly applicable to structural size timber sincestructural size timber invariably contains defects,which further reduces its strength. To take accountof this, timber was visually classified into one offour grades, namely 75, 65, 50 and 40, which ind-icated the percentage free from defects. The gradestress for structural size timber was finally obtainedby multiplying the grade designations expressed asa percentage (e.g. 75%, 65% etc.) by the basicstress for the timber.

With the introduction of BS 5268 the concept ofbasic stress was largely abandoned and a revisedprocedure for assessing the strength of timberadopted. From then on, the first step involved grad-ing structural size timber. Grading was still carriedout visually, although it was now common practiceto do this mechanically. The latter approach of-fered the advantage of greater economy in the useof timber since it took into account the density of

timber which significantly influences its strength.Mechanical stress grading is based on the fact

that there is a direct relationship between the mo-dulus of elasticity measured over a relatively shortspan, i.e. stiffness, and bending strength. The stiff-ness is assessed non-destructively by feeding indi-vidual pieces of timber through a series of rollerson a machine which automatically applies smalltransverse loads over short successive lengths andmeasures the deflections. These are compared withpermitted deflections appropriate to given stressgrades and the machine assesses the grade of thetimber over its entire length.

When BS 5268 was published in 1984 thenumbered grades (i.e. 75, 65, 50 and 40) werewithdrawn and replaced by two visual grades: Gen-eral Structural (GS) and Special Structural (SS) andfour machine grades: MGS, MSS, M75 and M50.The SS grade timber was used as the basis forstrength and modulus of elasticity determinationsby subjecting a large number of structural sizedspecimens to short-term load tests. The results wereused to obtain the fifth percentile stresses, definedas the value below which not more than 5% of testresults fell (Fig. 6.1). The fifth percentile valuesfor other grades of the same species were derivedusing grade relativity factors established from thesame series of tests. Finally, the grade stresses wereobtained by dividing the fifth percentile stressesby a reduction factor, which included adjustmentsfor a standard depth of specimen of 300 mm, dura-tion of load and a factor of safety. The two visualgrades are still referred to in the latest revision ofBS 5268 published in 2002. However, machinegraded timber is now graded directly to one of six-teen strength classes defined in BS EN 519, princi-pally on the basis of bending stress, mean modulusof elasticity and characteristic density, and markedaccordingly.

6.3 Grade stress andstrength class

Table 6.1 shows typical timber species/grade com-binations and associated grade stresses and moduliof elasticity. This information would enable thedesigner to determine the size of a timber membergiven the intensity and distribution of the loads tobe carried. However, it would mean that the con-tractor’s choice of material would be limited toone particular species/grade combination, whichcould be difficult to obtain. It would obviously bebetter if a range of species/grade combinations could

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281

Fig. 6.1 Frequency distribution curve for flexural strength of timber.

Table 6.1 Grade stresses for softwoods graded in accordance with BS 4978: for service classes 1and 2 (Table 10, BS 5268)

Standard name Grade Bending Tension Compression Shear Modulus of elasticityparallel parallel parallelto graina to graina Parallel Perpendicular to grain Mean Minimum

to grain to grainb

N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2

Redwood/whitewood SS 7.5 4.5 7.9 2.1 0.82 10 500 7 000(imported) GS 5.3 3.2 6.8 1.8 0.82 9 000 6 000British larch SS 7.5 4.5 7.9 2.1 0.82 10 500 7 000

GS 5.3 3.2 6.8 1.8 0.82 9 000 6 000British pine SS 6.8 4.1 7.5 2.1 0.82 10 500 7 000

GS 4.7 2.9 6.1 1.8 0.82 9 000 6 000British spruce SS 5.7 3.4 6.1 1.6 0.64 8 000 5 000

GS 4.1 2.5 5.2 1.4 0.64 6 500 4 500Douglas fir SS 6.2 3.7 6.6 2.4 0.88 11 000 7 000(British grown) GS 4.4 2.6 5.2 2.1 0.88 9 500 6 000Parana pine SS 9.0 5.4 9.5 2.4 1.03 11 000 7 500(imported) GS 6.4 3.8 8.1 2.2 1.03 9 500 6 000Pitch pine SS 10.5 6.3 11.0 3.2 1.16 13 500 9 000(Caribbean) GS 7.4 4.4 9.4 2.8 1.16 11 000 7 500Western red cedar SS 5.7 3.4 6.1 1.7 0.63 8 500 5 500(imported) GS 4.1 2.5 5.2 1.6 0.63 7 000 4 500Douglas fir-larch SS 7.5 4.5 7.9 2.4 0.85 11 000 7 500(Canada and USA) GS 5.3 3.2 6.8 2.2 0.85 10 000 6 500Hem-fir SS 7.5 4.5 7.9 1.9 0.68 11 000 7 500(Canada and USA) GS 5.3 3.2 6.8 1.7 0.68 9 000 6 000Spruce-pine-fir SS 7.5 4.5 7.9 1.8 0.68 10 000 6 500(Canada and USA) GS 5.3 3.2 6.8 1.6 0.68 8 500 5 500Sitka spruce SS 6.6 4.0 7.0 1.7 0.66 10 000 6 500(Canada) GS 4.7 2.8 6.0 1.5 0.66 8 000 5 500Western whitewoods SS 6.6 4.0 7.0 1.7 0.66 9 000 6 000(USA) GS 4.7 2.8 6.0 1.5 0.66 7 500 5 000Southern pine SS 9.6 5.8 10.2 2.5 0.98 12 500 8 500(USA) GS 6.8 4.1 8.7 2.2 0.98 10 500 7 000

Notes. a Stresses applicable to timber 300 mm deep (or wide): for other section sizes see 2.10.6 and 2.12.2 of BS 5268.b When the specifications specifically prohibit wane at bearing areas, the SS grade compression perpendicular to grain stress maybe multiplied by 1.33 and used for all grades.

Frequency ofoccurrence

5% test results Fifth percentile stress

Flexural strength

Mean strength

1.64 s.d.

Grade stress and strength class

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Table 6.2 Softwood combinations of species and visual grades which satisfy the requirements forvarious strength classes

Standard name Strength classes

C14 C16 C18 C22 C24 C27 C30

ImportedParana pine GS SSCaribbean pitch pine GS SSRedwood GS SSWhitewood GS SSWestern red cedar GS SS

Douglas fir-larch (Canada and USA) GS SSHem-fir (Canada and USA) GS SSSpruce-pine-fir (Canada and USA) GS SSSitka spruce (Canada) GS SSWestern whitewoods (USA) GS SSSouthern pine (USA) GS SS

British grownDouglas fir GS SSLarch GS SSBritish pine GS SSBritish spruce GS SS

be specified and the contractor could then selectthe most economical one. Such an approach formsthe basis of grouping timber species/grade com-binations with similar strength characteristics intostrength classes (Table 6.2).

In all there are sixteen strength classes, C14,C16, C18, C22, C24, TR26, C27, C30, C35, C40,D30, D35, D40, D50, D60 and D70, with C14having the lowest strength characteristics. Thestrength class designations indicate the bendingstrength of the timber. Strength classes C14 to C40and TR26 are for softwoods and D30 to D70 arefor hardwoods. Strength class TR26 is intendedfor use in the design of trussed rafters. The gradestresses and moduli of elasticity associated witheach strength class are shown in Table 6.3. In theUK structural timber design is normally based onstrength classes C16 to C27. These classes cover awide range of softwoods which display good struc-tural properties and are both plentiful and cheap.

6.4 Permissible stressesThe grade stresses given in Tables 6.1 and 6.3 werederived assuming particular conditions of service

and loading. In order to take account of the actualconditions that individual members will be sub-ject to during their design life, the grade stressesare multiplied by modification factors known asK-factors. The modified stresses are termed per-missible stresses.

BS 5268 lists over 80 K-factors. However, thefollowing subsections consider only those modifica-tion factors relevant to the design of simple flexuraland compression members, namely:

K2: Moisture content factorK3: Duration of loading factorK5: Notched ends factorK7: Depth factorK8: Load-sharing systems factorK12: Compression member stress factor.

6.4.1 MOISTURE CONTENT, K2The strength and stiffness of timber decreases withincreasing moisture content. This effect is takeninto account by assigning timber used for struc-tural work to a service class. BS 5628 recognisesthree service classes as follows:

Service class 1 is characterised by a moisture con-tent in the material corresponding to a temperature

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Table 6.3 Grade stresses and moduli of elasticity for various strength classes: for service classes 1and 2 (based on Tables 8 and 9, BS 5268)

Strength Bending Tension Compression Compression Shear Modulus Characteristic Averageclass parallel parallel parallel perpendicular parallel of elasticity density2 density2

to grain to grain to grain to grain1 to grain(σm,g,||) (σc,g,|| ) (σc,g,⊥ ) (τg ) Emean Emin ρk ρmean

N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 kg/m3 kg/m3

C14 4.1 2.5 5.2 2.1 1.6 0.60 6 800 4 600 290 350C16 5.3 3.2 6.8 2.2 1.7 0.67 8 800 5 800 310 370C18 5.8 3.5 7.1 2.2 1.7 0.67 9 100 6 000 320 380C22 6.8 4.1 7.5 2.3 1.7 0.71 9 700 6 500 340 410C24 7.5 4.5 7.9 2.4 1.9 0.71 10 800 7 200 350 420TR26 10.0 6.0 8.2 2.5 2.0 1.10 11 000 7 400 370 450C27 10.0 6.0 8.2 2.5 2.0 1.10 12 300 8 200 370 450C30 11.0 6.6 8.6 2.7 2.2 1.20 12 300 8 200 380 460C35 12.0 7.2 8.7 2.9 2.4 1.30 13 400 9 000 400 480C40 13.0 7.8 8.7 3.0 2.6 1.40 14 500 10 000 420 500D30 9.0 5.4 8.1 2.8 2.2 1.40 9 500 6 000 530 640D35 11.0 6.6 8.6 3.4 2.6 1.70 10 000 6 500 560 670D40 12.5 7.5 12.6 3.9 3.0 2.00 10 800 7 500 590 700D50 16.0 9.6 15.2 4.5 3.5 2.20 15 000 12 600 650 780D60 18.0 10.8 18.0 5.2 4.0 2.40 18 500 15 600 700 840D70 23.0 13.8 23.0 6.0 4.6 2.60 21 000 18 000 900 1 080

1 When the specification specifically prohibits wane at bearing areas, the higher values may be used.2 For the calculation of dead load, the average density should be used.

of 20°C and the relative humidity of the surround-ing air only exceeding 65% for a few weeks per year.Timbers used internally in a continuously heatedbuilding normally experience this environment. Insuch environments most timbers will attain an aver-age moisture content not exceeding 12%.

Service class 2 is characterised by a moisturecontent in the material corresponding to a tem-perature of 20°C and the relative humidity of thesurrounding air only exceeding 85% for a few weeksper year. Timbers used in covered buildings willnormally experience this environment. In such en-vironments most timbers will attain an averagemoisture content not exceeding 20%.

Service class 3, due to climatic conditions, ischaracterised by higher moisture contents thanservice class 2 and is applicable to timbers usedexternally and fully exposed.

The grade stresses and moduli of elasticity shownin Tables 6.1 and 6.3 apply to timber exposed toservice classes 1 and 2. According to clause 2.6.2of BS 5268 where service class 3 exists, the valuesin Tables 6.1 and 6.3 should be multiplied by amodification factor K2 given in Table 16 of BS 5268,reproduced here as Table 6.4. Clause 2.6.1 also

Table 6.4 Modification factor K2 by whichstresses and moduli for service classes 1 and 2should be multiplied to obtain stresses and moduliapplicable to service class 3 (Table 16, BS 5268)

Property K2

Bending parallel to grain 0.8Tension parallel to grain 0.8Compression parallel to grain 0.6Compression perpendicular to grain 0.6Shear parallel to grain 0.9Mean and minimum modulus of elasticity 0.8

Permissible stresses

notes that because it is difficult to dry thick timber,service class 3 stresses and moduli should be usedfor solid timber members more than 100 mm thick,unless they have been specially dried.

6.4.2 DURATION OF LOADING, K3The stresses given in Tables 6.1 and 6.3 apply tolong-term loading. Where the applied loads willact for shorter durations e.g. snow and wind, the

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Table 6.5 Modification factor K3 for duration of loading (Table 17, BS 5268)

Duration of loading Value of K3

Long term (e.g. dead + permanent imposeda) 1.00Medium term (e.g. dead + snow, dead + temporary imposed) 1.25Short term (e.g. dead + imposed + windb, dead + imposed + snow + windb) 1.50Very short term (e.g. dead + imposed + windc) 1.75

Notes. a For uniformly distributed imposed floor loads K3 = 1.00 except for type C3 occupancy (Table 1, BS 6399: Part 1:1996) where for foot traffic on corridors, hallways, landings and stairways only, K3 may be assumed to be 1.5.b For wind where the largest diagonal dimension of the loaded area a, as defined in BS 6399: Part 2, exceeds 50 m.c For wind, very short-term category applies to classes A and B (3 s or 5 s gust) as defined in CP3 : Chapter V : Part 2 or,where the largest diagonal dimension of the loaded area a, as defined in BS 6399: Part 2, does not exceed 50 m.

Fig. 6.2 Notched beams: (a) beam with notch ontop edge; (b) beam with notch on underside (Fig. 2,BS 5268).

he

he

h

h

a

(a)

(b)

1. For a notch on the top edge (Fig. 6.2(a)):

K

h5

e e

e2

( )

=− +h a ah

hfor a ≤ he (6.1)

K5 = 1.0 for a > he (6.2)

2. For a notch on the underside (Fig. 6.2(b)):

K

hh5

e = (6.3)

Clause 2.10.4 of BS 5268 also notes that theeffective depth, he, should not be less than 0.5 h,i.e. K5 ≥ 0.5.

6.4.4 DEPTH FACTOR, K7The grade bending stresses given in Table 6.3 onlyapply to timber sections having a depth h of 300mm. For other depths of beams, the grade bendingstresses are multiplied by the depth factor K7,defined in clause 2.10.6 of BS 5268 as follows:

K7 = 1.17 for solid beams having a depth≤ 72 mm

K

h7

0.11

300

=

for solid beams with

72 mm < h < 300 mm (6.4)

K

hh7

2

2

0.81( 92300)( 56800)

=+

+ for solid beams

with h > 300 mm

6.4.5 LOAD-SHARING SYSTEMS, K8The grade stresses given in Tables 6.1 and 6.3apply to individual members, e.g. isolated beamsand columns, rather than assemblies. When four

grade stresses can be increased. Table 17 of BS5268, reproduced here as Table 6.5, gives the modi-fication factor K3 by which these values should bemultiplied for various load combinations.

6.4.3 NOTCHED ENDS, K5Notches at the ends of flexural members will re-sult in high shear concentrations which may causestructural failure and must, therefore, be taken intoaccount during design (Fig. 6.2).

In notched members the grade shear stressesparallel to the grain (Tables 6.1 and 6.3) are mul-tiplied by a modification factor K5 calculated asfollows:

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285

or more members such as rafters, joists or wallstuds, spaced a maximum of 610 mm centre tocentre act together to resist a common load, thegrade stress should be multiplied by a load-sharingfactor K8 which has a value of 1.1 (clause 2.9,BS 5268).

6.4.6 COMPRESSION MEMBERS, K12The grade compression stresses parallel to the graingiven in Tables 6.1 and 6.3 are used to design strutsand columns. These values apply to compressionmembers with slenderness ratios less than 5 whichwould fail by crushing. Where the slenderness ratioof the member is equal to or greater than 5 thegrade stresses should be multiplied by the modifi-cation factor K12 given in Table 22 of BS 5268,reproduced here as Table 6.6. Alternatively Appen-dix B of BS 5268 gives a formula for K12 whichcould be used. This is based on the Perry-Robertsonequation which is also used to model the behaviourof steel compression members (section 4.9). Thefactor K12 takes into account the tendency of themember to fail by buckling and allows for imper-fections such as out of straightness and accidentalload eccentricities.

The factor K12 is based on the minimummodulus of elasticity, Emin, irrespective of whetherthe compression member acts alone or forms partof a load-sharing system and the compression stress,σc,||, is given by:

σc,|| = σc,g,||K3 (6.5)

6.5 Timber designHaving discussed some of the more general as-pects, the following sections will consider in detailthe design of:

1. flexural members2. compression members3. stud walling.

6.6 SymbolsFor the purposes of this chapter, the following sym-bols have been used. These have largely been takenfrom BS 5268.

GEOMETRICAL PROPERTIESb breadth of beamh depth of beamA total cross-sectional area

i radius of gyrationI second moment of areaZ elastic modulus

BENDINGL effective spanM design momentMR moment of resistanceσm,a,|| applied bending stress parallel to grainσm,g,|| grade bending stress parallel to grainσm,adm,|| permissible bending stress parallel to

grain

DEFLECTIONδt total deflectionδm bending deflectionδv shear deflectionδp permissible deflectionE modulus of elasticityEmean mean modulus of elasticityEmin minimum modulus of elasticityG shear modulus

SHEARFv design shear forceτa applied shear stress parallel to grainτg grade shear stress parallel to grainτadm permissible shear stress parallel to grain

BEARINGF bearing forcelb length of bearingσc,a,⊥ applied compression stress

perpendicular to grainσc,g,⊥ grade compression stress perpendicular

to grainσc,adm,⊥ permissible bending stress perpendicular

to grain

COMPRESSIONLe effective length of a columnλ slenderness ratioN axial loadσc,a,|| applied compression stress parallel

to grainσc,g,|| grade compression stress parallel

to grainσc,adm,|| permissible compression stress parallel

to grainσc,|| compression stress = σc,g,||K3

σe Euler critical stress

Symbols

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286

Tab

le 6

.6M

odifi

cati

on f

acto

r K

12 f

or c

ompr

essi

on m

embe

rs (

Tab

le 2

2, B

S 5

268)

E/σ

c,||

Val

ue o

f K

12

Val

ues

of s

lend

erne

ss r

atio

λ (

= L

e/i)

< 5

510

2030

4050

6070

8090

100

120

140

160

180

200

220

240

250

Equ

ival

ent

Le/b

(fo

r re

ctan

gula

r se

ctio

ns)

< 1.

41.

42.

95.

88.

711

.614

.517

.320

.223

.126

.028

.934

.740

.546

.252

.057

.863

.669

.472

.3

400

1.00

00.

975

0.95

10.

896

0.82

70.

735

0.62

10.

506

0.40

80.

330

0.27

10.

225

0.16

20.

121

0.09

40.

075

0.06

10.

051

0.04

30.

040

500

1.00

00.

975

0.95

10.

899

0.83

70.

759

0.66

40.

562

0.46

60.

385

0.32

00.

269

0.19

50.

148

0.11

50.

092

0.07

60.

063

0.05

30.

049

600

1.00

00.

975

0.95

10.

901

0.84

30.

774

0.69

20.

601

0.51

10.

430

0.36

30.

307

0.22

60.

172

0.13

50.

109

0.08

90.

074

0.06

30.

058

700

1.00

00.

975

0.95

10.

902

0.84

80.

784

0.71

10.

629

0.54

50.

467

0.39

90.

341

0.25

40.

195

0.15

40.

124

0.10

20.

085

0.07

20.

067

800

1.00

00.

975

0.95

20.

903

0.85

10.

792

0.72

40.

649

0.57

20.

497

0.43

00.

371

0.28

00.

217

0.17

20.

139

0.11

50.

096

0.08

20.

076

900

1.00

00.

976

0.95

20.

904

0.85

30.

797

0.73

40.

665

0.59

30.

522

0.45

60.

397

0.30

40.

237

0.18

80.

153

0.12

70.

106

0.09

10.

084

1000

1.00

00.

976

0.95

20.

904

0.85

50.

801

0.74

20.

677

0.60

90.

542

0.47

80.

420

0.32

50.

255

0.20

40.

167

0.13

80.

116

0.09

90.

092

1100

1.00

00.

976

0.95

20.

905

0.85

60.

804

0.74

80.

687

0.62

30.

559

0.49

70.

440

0.34

40.

272

0.21

90.

179

0.14

90.

126

0.10

70.

100

1200

1.00

00.

976

0.95

20.

905

0.85

70.

807

0.75

30.

695

0.63

40.

573

0.51

30.

457

0.36

20.

288

0.23

30.

192

0.16

00.

135

0.11

60.

107

1300

1.00

00.

976

0.95

20.

905

0.85

80.

809

0.75

70.

701

0.64

30.

584

0.52

70.

472

0.37

80.

303

0.24

70.

203

0.17

00.

144

0.12

30.

115

1400

1.00

00.

976

0.95

20.

906

0.85

90.

811

0.76

00.

707

0.65

10.

595

0.53

90.

486

0.39

20.

317

0.25

90.

214

0.18

00.

153

0.13

10.

122

1500

1.00

00.

976

0.95

20.

906

0.86

00.

813

0.76

30.

712

0.65

80.

603

0.55

00.

498

0.40

50.

330

0.27

10.

225

0.18

90.

161

0.13

80.

129

1600

1.00

00.

976

0.95

20.

906

0.86

10.

814

0.76

60.

716

0.66

40.

611

0.55

90.

508

0.41

70.

342

0.28

20.

235

0.19

80.

169

0.14

50.

135

1700

1.00

00.

976

0.95

20.

906

0.86

10.

815

0.76

80.

719

0.66

90.

618

0.56

70.

518

0.42

80.

353

0.29

20.

245

0.20

70.

177

0.15

20.

142

1800

1.00

00.

976

0.95

20.

906

0.86

20.

816

0.77

00.

722

0.67

30.

624

0.57

40.

526

0.43

80.

363

0.30

20.

254

0.21

50.

184

0.15

90.

148

1900

1.00

00.

976

0.95

20.

907

0.86

20.

817

0.77

20.

725

0.67

70.

629

0.58

10.

534

0.44

70.

373

0.31

20.

262

0.22

30.

191

0.16

50.

154

2000

1.00

00.

976

0.95

20.

907

0.86

30.

818

0.77

30.

728

0.68

10.

634

0.58

70.

541

0.45

50.

382

0.32

00.

271

0.23

00.

198

0.17

20.

160

9780415467193_C06 9/3/09, 2:14 PM286

287

Fig. 6.3 Effective span of simply supported beams.

BearingClear span

Effective span

Bearing e.g. wall plate

Flexural members

6.7 Flexural membersBeams, rafters and joists are examples of flexuralmembers. All calculations relating to their designare based on the effective span and principally in-volves consideration of the following aspects whichare discussed below:

1. bending2. deflection3. lateral buckling4. shear5. bearing.

Generally, for medium-span beams the design pro-cess follows the sequence indicated above. However,deflection is usually critical for long-span beams andshear for heavily loaded short-span beams.

6.7.1 EFFECTIVE SPANAccording to clause 2.10.3 of BS 5268, for simplysupported beams, the effective span is normallytaken as the distance between the centres of bear-ings (Fig. 6.3).

6.7.2 BENDINGIf flexural members are not to fail in bending, thedesign moment, M, must not exceed the momentof resistance, MR

M ≤ MR (6.6)

The design moment is a function of the appliedloads. The moment of resistance for a beam can bederived from the theory of bending (equation 2.5,Chapter 2) and is given by

MR = σm,adm,||Zxx (6.7)

whereσm,adm,|| permissible bending stress parallel to grainZxx section modulus

For rectangular sections Zxx =

bd2

6(Fig. 6.4)

whereb breadth of sectiond depth of section

The permissible bending stress is calculated bymultiplying the grade bending stress, σm,g,||, by anyrelevant K-factors:

σm,adm,|| = σm,g,||K2K3K7K8 (as appropriate)(6.8)

For a given design moment the minimum requiredsection modulus, Zxx req, can be calculated usingequation 6.9, obtained by combining equations 6.6and 6.7:

Zxx req ≥

Mσm,adm,||

(6.9)

A suitable timber section can then be selectedfrom Tables NA.2, NA.3 and NA.4 of BS EN336: Structural timber. Sizes permitted deviations.These tables give the commonly available sizes of,respectively, sawn timber, timber machined on thewidth and timber machined on all four sides. Ta-ble NA.2 is reproduced here as Table 6.7. Table 6.8is an expanded version which includes a number ofuseful section properties to aid design. Finally, thechosen section should be checked for deflection,lateral buckling, shear and bearing to assess its suit-ability as discussed below.

x x

b

d

Fig. 6.4 Section modulus.

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288

Table 6.7 Commonly available target sizes of sawn softwoodstructural timber (based on Table NA.2, BS EN 336)

Thickness (mm) Width (to tolerance class 1) (mm)(to toleranceclass 1) 75 100 125 150 175 200 225 250 300

22 x38 x x x x x47 x x x x x x x x x63 x x x x75 x x x x x x x100 x x x x x x150 x x300 x x

Note 1 Timber in other sizes is available to orderNote 2 In the UK the above sizes are commonly available in strength classes

C16 and C24

Table 6.8 Geometrical properties of sawn softwoods

Customary Area Section Modulus Second of moment area Radius of gyrationtarget size*

About x–x About y–y About x–x About y–y About x–x About y–ymm 103mm2 103mm3 103mm3 106mm4 106mm4 mm mm

22 × 100 2.20 36.6 8.1 1.83 0.089 28.9 6.35

38 × 100 3.80 63.3 24.1 3.17 0.457 28.9 11.038 × 150 5.70 143 36.1 10.7 0.686 43.3 11.038 × 175 6.54 194 42.1 17.0 0.800 50.5 11.038 × 200 7.60 253 48.1 25.3 0.915 57.7 11.038 × 225 8.55 321 54.2 36.1 1.03 65.0 11.0

47 × 75 3.53 44.1 27.6 1.65 0.649 21.7 13.647 × 100 4.70 78.3 36.8 3.92 0.865 28.9 13.647 × 125 5.88 122 46.0 7.65 1.08 36.1 13.647 × 150 7.05 176 55.2 13.2 1.30 43.3 13.647 × 175 8.23 240 64.4 21.0 1.51 50.5 13.647 × 200 9.40 313 73.6 31.3 1.73 57.7 13.647 × 225 10.6 397 82.8 44.6 1.95 65.0 13.647 × 250 11.8 490 92.0 61.2 2.16 72.2 13.647 × 300 14.1 705 110 106 2.60 86.6 13.6

63 × 150 9.45 236 99.2 17.7 3.13 43.3 18.263 × 175 11.0 322 116 28.1 3.65 50.5 18.263 × 200 12.6 420 132 42.0 4.17 57.7 18.263 × 225 14.2 532 149 59.8 4.69 65.0 18.2

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289


Customary Area Section Modulus Second moment of area Radius of gyrationtarget size*

About x–x About y–y About x–x About y–y About x–x About y–ymm 103mm2 103mm3 103mm3 106mm4 106mm4 mm mm

Flexural members

75 × 100 7.50 125 93.8 6.25 3.52 28.9 21.775 × 150 11.3 281 141 21.1 5.27 43.3 21.775 × 175 13.1 383 164 33.5 6.15 50.5 21.775 × 200 15.0 500 188 50.0 7.03 57.7 21.775 × 225 16.9 633 211 71.2 7.91 65.0 21.775 × 250 18.8 781 234 97.7 8.79 72.2 21.775 × 300 22.5 1130 281 169 10.5 86.6 21.7

100 × 100 10.0 167 167 8.33 8.33 28.9 28.9100 × 150 15.0 375 250 28.1 12.5 43.3 28.9100 × 200 20.0 667 333 66.7 16.7 57.7 28.9100 × 225 22.5 844 375 94.9 18.8 65.0 28.9100 × 250 25.0 1010 417 130 20.8 72.2 28.9100 × 300 30.0 1500 500 225 25.0 86.6 28.9

150 × 150 20.0 563 563 42.2 42.2 43.3 43.3150 × 300 30.0 2250 1130 338 84.4 86.6 43.3

300 × 300 90.0 4500 4500 675 675 86.6 86.6

Note. * Desired size of timber measured at 20% moisture content

6.7.3 DEFLECTIONExcessive deflection of flexural members mayresult in damage to surfacing materials, ceilings,partitions and finishes, and to the functional needsas well as aesthetic requirements.

Clause 2.10.7 of BS 5268 recommends thatgenerally such damage can be avoided if the totaldeflection, δt, of the member when fully loadeddoes not exceed the permissible deflection, δp:

δt ≤ δp (6.10)

The permissible deflection is generally given by

δp = 0.003 × span (6.11)

but for longer-span domestic floor joists, i.e. spansover 4.67 m, should not exceed 14 mm:

δp ≤ 14 mm (6.12)

The total deflection, δt, is the summation of thebending deflection, δm, plus the shear deflection,δv:

δt = δm + δv (6.13)

Table 6.9 gives the bending and shear deflectionformulae for some common loading cases forbeams of rectangular cross-section. The formulaehave been derived by assuming that the shearmodulus is equal to one-sixteenth of the permis-sible modulus of elasticity in accordance with clause2.7 of BS 5268.

For solid timber members acting alone the de-flections should be calculated using the minimummodulus of elasticity, but for load-sharing sys-tems the deflections should be based on the meanmodulus of elasticity.

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290

Table 6.9 Bending and shear deflectionsassuming G = E/16

Load distributionand supports

6.7.4 LATERAL BUCKLINGIf flexural members are not effectively laterallyrestrained, it is possible for the member to twistsideways before developing its full flexural strength(Fig. 6.5 ), thereby causing it to fail in bending,shear or deflection. This phenomenon is calledlateral buckling and can be avoided by ensuringthat the depth to breadth ratios given in Table 6.10are complied with.

6.7.5 SHEARIf flexural members are not to fail in shear, theapplied shear stress parallel to the grain, τa, shouldnot exceed the permissible shear stress, τadm:

τa ≤ τadm (6.14)

L

w C

L/2 L/2C

W

L

WC

W

a a

L/2 L/2C

W

L E

W

LE

w

Bending

Deflection atCentre C or end E

Table 6.10 Maximum depth to breadth ratios (Table 19, BS 5268)

Degree of lateral support Maximum depthto breadth ratio

No lateral support 2Ends held in position 3Ends held in position and member held in line, as by purlins or 4

tie-rods at centres not more than 30 times the breadth of the memberEnds held in position and compression edge held in line, 5

as by direct connection of sheathing, deck or joistsEnds held in position and compression edge held in line, as by direct 6

connection of sheathing, deck or joists, together with adequate bridgingor blocking spaced at intervals not exceeding six times the depth

Ends held in position and both edges held firmly in line 7

L

w C

Shear

WLEI

3

48

245

×WLEA

WaEI

L a2 2

8 6 −

965

×WaEA

5384

4

×wLEI

125

2

×wLEA

wLEI

4

384

125

2

×wLEA

WLEI

3

192

245

×WLEA

wLEI

4

8

485

2

×wLEA

WLEI

3

3

965

×WLEA

Fig. 6.5 Lateral buckling.

Original position ofbeam shown dotted

Lateraldisplacement

Lateral displacementSpan

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291

For a beam with a rectangular cross-section, themaximum applied shear stress occurs at the neutralaxis and is given by:

τa

v32

= FA

(6.15)

whereFv applied maximum vertical shear forceA cross-sectional area

The permissible shear stress is given by

τadm = τgK2K3K5K8 (as appropriate) (6.16)

where τg is the grade shear stress parallel to thegrain (Tables 6.1 and 6.3).

6.7.6 BEARING PERPENDICULAR TO GRAINBearing failure may arise in flexural members whichare supported at their ends on narrow beams orwall plates. Such failures can be avoided by en-suring that the applied bearing stress, σc,a,⊥, neverexceeds the permissible compression stress perpen-dicular to the grain, σc,adm,⊥:

σc,a,⊥ ≤ σc,adm,⊥ (6.17)

The applied bearing stress is given by

σc,a,⊥ =

Fblb

(6.18)

whereF bearing force (usually maximum reaction)b breadth of sectionlb bearing length.

Wane

Fig. 6.6 Wane.

Flexural members

The permissible compression stress is obtainedby multiplying the grade compression stress per-pendicular to the grain, σc,g,⊥, by the K-factors formoisture content (K2), load duration (K3) and loadsharing (K8) as appropriate:

σc,adm,⊥ = σc,g,⊥K2K 3K 8 (6.19)

It should be noted that the grade compressionstresses perpendicular to the grain given in Tables6.1 and 6.3 apply to (i) bearings of any length atthe ends of members and (ii) bearings 150 mmor more in length at any position. Moreover, twovalues for the grade compression stress perpendicu-lar to the grain are given for each strength class(Table 6.3). The lower value takes into account theamount of wane which is permitted within eachstress grade (Fig. 6.6). If, however, the specificationprohibits wane from occurring at bearing areas thehigher value may be used.

EFFECTIVE SPANDistance between centres of bearing (l ) = 3000 mm

150 2850 150

W = 10 kN

Example 6.1 Design of a timber beam (BS 5268)A timber beam with a clear span of 2.85 m supports a uniformly distributed load of 10 kN including self-weight ofbeam. Determine a suitable section for the beam using timber of strength class C16 under service class 1. Assumethat the bearing length is 150 mm and that the ends of the beam are held in position and compression edge held inline.

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292


GRADE STRESS AND MODULUS OF ELASTICITY FOR C16Values in N/mm2 are as follows

Bending parallel Shear parallel Compression Modulus ofto grain to grain perpendicular to grain elasticityσm,g,|| τg σc,g,⊥ Emin

5.3 0.67 1.7 5800

MODIFICATION FACTORSK2, moisture content factor does not apply since the beam is subject to service class 1K3, duration of loading factor = 1.0K8, load sharing factor, does not apply since there is only a single beam

K7, depth factor =

3000.11

h

Assume h = 250, K7 = 1.020

BENDING

M

W l

8

10 38

3.75 kN m= = =×

σm,adm,||(assuming h = 250) = σm,g,||K3K7 = 5.3 × 1.0 × 1.020 = 5.406 N/mm2

Zxxreq ≥

Mσm,adm,

63.75 105.406||

=×

= 694 × 103 mm3

DEFLECTIONPermissible deflection (δp) = 0.003 × spanThe deflection due to shear (δs) is likely to be insignificant in comparison to the bending deflection (δb) and may beignored in order to make a first estimate of the total deflection (δt):

δt(ignoring shear deflection) =

5384

3

min xx

WlE I

(Table 6.9)

=

× ×× ×

5 10 3000384 5800

4 3

xx

I

Since δp ≥ δt

0.003 × 3000 ≥

5 10 3000384 5800

4 3

xx

× ×× ×

I

Ixxreq ≥ 67.3 × 106 mm4

From Table 6.8, section 75 × 250 providesZxx = 781 × 103 mm3 Ixx = 97.7 × 106 mm4 A = 18.8 × 103 mm2

Hence total deflection including shear deflection can now be calculated and is given by

5384

125

5 10 3000

384 5800 97.7 10

12 10 30005 5800 18.8 10

3

min xx min

4 3

6

4

3

WlE I

WlE A

+ =× ×

× × ×+

× ×× × ×

= 6.2 mm + 0.7 mm = 6.9 mm ≤ δp = 0.003 × 3000 = 9 mm

Therefore a beam with a 75 × 250 section is adequate for bending and deflection.

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293

LATERAL BUCKLING

Permissible

db

= 5 (Table 6.10)

Actual

db

25075

3.3= = < permissible

Hence the section is adequate for lateral buckling.

SHEARPermissible shear stress is

τadm = τgK 3 = 0.67 × 1.0 = 0.67 N/mm2

Maximum shear force is

F v =

W2

=

×10 102

3

= 5 × 103 N

Maximum shear stress at neutral axis is

τa =

32

32

5 1018.8 10

v3

3

FA

= ××

× = 0.4 N/mm2 < permissible

Therefore the section is adequate in shear.

BEARINGPermissible bearing stress is

σc,adm,⊥ = σc,g,⊥K3 = 1.7 × 1.0 = 1.7 N/mm2

End reaction, F, is

W2

=

×10 102

3

= 5 × 103 N

σc,a,⊥ =

Fblb

=××

5 1075 150

3

= 0.44 N/mm2 < permissible

Therefore the section is adequate in bearing. Since all the checks are satisfactory, use 75 mm × 250 mm sawn C16 beam.

Floor decktongue and grooveboarding

Plasterboardceiling

400 mm 400 mm

Example 6.2 Design of timber floor joists (BS 5268)Design the timber floor joist for a domestic dwelling using timber of strength class C18 given that:

a) the joists are spaced at 400 mm centres;b) the floor has an effective span of 3.8 m;c) the flooring is tongue and groove boarding with a self-weight of 0.1 kN/m2;d) the ceiling is of plasterboard with a self weight of 0.2 kN/m2.


Flexural members

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294

Example 6.2 continuedDESIGN LOADING

Tongue and groove boarding = 0.10 kN/m2

Ceiling = 0.20 kN/m2

Joists (say) = 0.10 kN/m2

Imposed floor load for domestic dwelling (Table 2.2) = 1.50 kN/m2

Total load = 1.90 kN/m2

Uniformly distributed load/joist (W ) is

W = joist spacing × effective span × load

= 0.4 × 3.8 × 1.9 = 2.9 kN

GRADE STRESSES AND MODULUS OF ELASTICITY FOR C18Values in N/mm2 are as follows

Bending parallel Compression Shear parallel Modulus ofto grain perpendicular to grain to grain elasticityσm,g,|| σc,g,⊥ τg Emean

5.8 1.7 0.67 9100

MODIFICATION FACTORSK2, moisture content factor does not apply since joists are exposed to service class 2K3, duration of loading = 1.0K8, load-sharing system = 1.1

K7, depth factor =

3000.11

h

whereh = 225, K7 = 1.032h = 200, K7 = 1.046h = 175, K7 = 1.061

BENDING

Bending moment (M) =

Wl8

kN m

. . . = =

2 9 3 88

1 4×

σm,adm,||(ignoring K7) = σm,g,||K3K8 = 5.8 × 1.0 × 1.1 = 6.38 N/mm2

Zxxreq ≥

Mσ

=×

m,adm,

||

.

.1 4 10

6 38

6

= 219 × 103 mm3

From Table 6.8 a 47 × 200 mm joist would be suitable (Zxx = 313 × 103 mm3, Ixx = 31.3 × 106 mm4, A = 9.4 × 103 mm2).Hence K7 = 1.046. Therefore

Zxxreq =

219 103×1 046.

= 209 × 103 mm3 < provided OK

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295

DEFLECTIONPermissible deflection = 0.003 × span

= 0.003 × 3800 = 11.4 mm

Total deflection (δt) = bending deflection (δm) + shear deflection (δv)

= +

5384

12

5

3

mean xx mean

WlE I

WlE A

=

× × × ×× × × ×

+× × × ×× × × ×

5 2.9 10 (3.8 10

384 9.1 10 31.3 10

12 2.9 10 3.8 105 9.1 10 9.4 10

3 3 3

3 6

3 3

3 3

)

= 7.3 mm + 0.3 mm = 7.6 mm < permissible

Therefore 47 mm × 200 mm joist is adequate in bending and deflection.

LATERAL BUCKLING

Permissible

db

= 5 (Table 6.10)

Actual

db

20047

= .= 4 3 < permissible

Therefore joist is satisfactory in lateral buckling.

SHEARPermissible shear stress is

τadm = τgK3 K8 = 0.67 × 1.0 × 1.1 = 0.737 N/mm2


Fv =

W2

.=

2 9 102

3× = 1.45 × 103 N

Maximum shear stress at neutral axis is

τa =

32

32

1.45 109.4 10

v3

3

FA

= ×××

= 0.23 N/mm2 < permissible

Therefore joist is adequate in shear.

BEARINGPermissible compression stress perpendicular to grain is

σc,adm,⊥ = σc,g,⊥K3K8 = 1.7 × 1.0 × 1.1 = 1.87 N/mm2

Maximum end reaction is

F =

W2

.=

×2 9 102

3

= 1.45 × 103 N

Assuming that the floor joists span on to 100 mm wide wall plates the bearing stress is given by

σc,a,⊥ =

Fbl b

=

××

1 45 1047 100

3. = 0.31 N/mm2 < permissible

Therefore joist is adequate in bearing.


Flexural members

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296

Example 6.4 Analysis of a timber roof (BS 5268)A flat roof spanning 4.5 m is constructed using timber joists of grade GS whitewood with a section size of 47 mm ×225 mm and spaced at 450 mm centres. The total dead load due to the roof covering and ceiling including the self-weight of the joists is 1 kN/m2. Calculate the maximum imposed load the roof can carry assuming that the durationof loading is (a) long term (b) medium term.

DESIGN LOADINGDead load = 1 kN/m2

Live load = q kN/m2

Uniformly distributed load/joist, W, is

W = joist spacing × effective span × (dead + live) = 0.45 × 4.5 (1 + q)

CHECK ASSUMED SELF-WEIGHT OF JOISTSFrom Table 6.3, the average density of timber of strength class C18 is 380 kg/m3. Hence, self-weight of the joists is

(47 200 10 ) 380 kg/m × × × × ×−3 3 39 81 10

0 4

.

.

−

= 0.088 kN/m2 < 0.10 kN/m2 (assumed)

Since all the checks are satisfactory use 47 mm × 200 mm C18 sawn floor joists.


Example 6.3 Design of a notched floor joist (BS 5268)The joists in Example 6.2 are to be notched at the bearings with a 75 m deep notch as shown below. Check that thenotched section is still adequate.

The presence of the notch affects only the shear stresses in the joists. For a notched member the permissible shearstress is given by

τadm = τgK3K5K8

where

K

hh5

e 125200

= = = 0.625 > min. (= 0.5)

Hence

τadm = 0.67 × 1.0 × 0.625 × 1.1 = 0.46 N/mm2

Applied shear parallel to grain, τa (from above) is

0.23 N/mm2 < permissible

Therefore the 47 mm × 200 mm sawn joists are also adequate when notched with a 75 mm deep bottom edge notchat the bearing.

Notch = 75 mm

he = 125 mmh = 200 mm

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297

Example 6.4 continuedGRADE STRESSES AND MODULUS OF ELASTICITYGrade GS whitewood timber belongs to strength class C16 (Table 6.2). Values in N/mm2 are as follows:

Bending parallel Compression Shear parallel Modulus ofto grain perpendicular to grain to grain elasticityσm,g,|| σc,g,⊥ τg Emean

5.3 1.7 0.67 8800

MODIFICATION FACTORSK3, duration of loading (Table 6.5) = 1.0 (long term) = 1.25 (medium term)K8, load-sharing system = 1.1

K7, depth factor =

3000.11

h

where h = 225, K7 = 1.032

GEOMETRICAL PROPERTIESFrom Table 6.8, 47 × 225 section provides:

Cross-sectional area, A = 10.6 × 103 mm2

Elastic modulus about x–x, Zxx = 397 × 103 mm3

Second moment of area about x–x, Ixx = 44.6 × 106 mm4

BENDINGLong termPermissible bending stress parallel to grain is

σm,adm,|| = σm,g,||K3K7K 8 = 5.3 × 1.0 × 1.032 × 1.1 = 6.02 N/mm2

Moment of resistance, MR, is

MR = σm,adm,||Z xx = 6.02 × 397 × 103 × 10−6 = 2.39 kN m

Design moment, M =

Wl8

= 0.45 × 4.5(1 + q)

4 58.

= 1.139(1 + q)

Equating MR = M,

2.39 = 1.139 (1 + q) ⇒ q = 1.09 kN/m2

Medium termFrom above

σm,adm,|| = 6.02K3(medium term) = 6.02 × 1.25 = 7.52 N/mm2

MR = 7.52 × 397 × 103 × 10−6 = 2.98 kNm

Equating MR = M,

2.98 = 1.139(1 + q) ⇒ q = 1.62 kN/m2

DEFLECTIONMaximum total deflection = bending deflection (δm) + shear deflection (δv)

0.003

5384

12

5

3

mean xx mean

LWLE I

WLE A

= +

Flexural members

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298

1. slenderness ratio2. axial compressive stress3. permissible compressive stress.

The following subsections consider these more gen-eral aspects before describing in detail the design ofthe above two categories of compression members.

6.8.1 SLENDERNESS RATIOThe load-carrying capacity of compression mem-bers is a function of the slenderness ratio, λ, whichis given by

6.8 Design of compressionmembers

Struts and columns are examples of compressionmembers. For design purposes BS 5268 dividescompression members into two categories (1)members subject to axial compression only and(2) members subject to combined bending and axialcompression.

The principal considerations in the design ofcompression members are:


Permissible shear parallel to grain isτadm = τg K3K8 = 0.67 × 1.0 × 1.1 = 0.737 N/mm2

Maximum shear force Fv =

23

τadm A (equation 6.15)

=

23

. . × × ×

0 737 10 6 103 × 10−3 = 5.2 kN

Total load per joist = 2Fv = 10.4 kN

Load per unit area =

10 40 45 4 5

.. .×

= 5.13 kN/m2

Henceq = 5.13 − 1 = 4.13 kN/m2 (long term)

andq = 5.43 kN/mm2 (medium term, K3 = 1.25)

Hence the safe long-term imposed load that the roof can support is 1.09 kN/m2 (bending critical) and the safemedium-term imposed load is 1.12 kN/m2 (deflection critical).

0.003

5 (4.5 10384 8800 44.6 10

12

5 8800 10.6 10

3

6 3=

× ×× × ×

+× × ×

W)2

= 6.975 × 10−7 WW = 4300 N per joist

Load per unit area is

Wjoist spacing span×

.

. .=

×4 3

0 45 4 5 = 2.12 kN/m2

Henceq = 2.12 − dead load = 2.12 − 1 = 1.12 kN/m2

SHEAR

W2

W2

W

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299

Effective lengthActual length

λ =

Lie (6.20)

whereLe effective lengthi radius of gyration

According to clause 2.11.4 of BS 5268, the slen-derness ratio should not exceed 180 for compres-sion members carrying dead and imposed loadsother than loads resulting from wind in which casea slenderness ratio of 250 may be acceptable.

The radius of gyration, i, is given by

i = I A/ (6.21)

whereI moment of inertiaA cross-section area.

For rectangular sections

i = b/ 12 (6.22)

where b is the least lateral dimension.The effective length, Le, of a column is obtained

by multiplying the actual length, L, by a coefficienttaken from Table 6.11 which is a function of thefixity at the column ends.

Le = L × coefficient (6.23)

In Table 6.11 end condition (a) models the case ofa column with both ends fully fixed and no relativehorizontal motion possible between the columnends. End condition (c) models the case of a pin-ended column with no relative horizontal motionpossible between column ends. End condition (e)models the case of a column with one end fullyfixed and the other free. Figure 6.7 illustrates allfive combinations of end fixities.

Table 6.11 Effective length of compression members (Table 21, BS 5268)

End conditions

(Le /L)

(a) Restrained at both ends in position and in direction 0.7(b) Restrained at both ends in position and one end in direction 0.85(c) Restrained at both ends in position but not in direction 1.0(d) Restrained at one end in position and in direction and at 1.5

the other end in direction but not in position(e) Restrained at one end in position and in direction and free 2.0

at the other end

6.8.2 AXIAL COMPRESSIVE STRESSThe axial compressive stress is given by

σc,a,|| =

FA

(6.24)

whereF axial loadA cross-sectional area.

6.8.3 PERMISSIBLE COMPRESSIVE STRESSAccording to clause 2.11.5 of BS 5268, for com-pression members with slenderness ratios of lessthan 5, the permissible compressive stress should betaken as the grade compression stress parallel to thegrain, σc,g,||, modified as appropriate for moisturecontent, duration of loading and load sharing:

σc,adm,|| = σc,g,||K2K3K8 for λ < 5 (6.25)

For compression members with slenderness ratiosequal to or greater than 5, the permissible com-pressive stress is obtained in the same way butshould additionally be modified by the factor K12

σc,adm,|| = σc,g,||K2K3K8K12 for λ ≥ 5 (6.26)

6.8.4 MEMBER DESIGNHaving discussed these common aspects it is nowpossible to describe in detail the design of com-pression members. As pointed out earlier, BS 5268distinguishes between two categories of members,that is, those subject to (a) axial compression onlyand (b) axial compression and bending.

6.8.4.1 Members subject to axialcompression only

This category of compression member is designedso that the applied compressive stress, σc,a,||, doesnot exceed the permissible compressive stress par-allel to the grain, σc,adm,||:


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300

Fig. 6.8 Bending in timber columns.

δ1 > δδ δ1

MM

F

M M

F

Fig. 6.7 End conditions.

(a)

L e =

2.0

L

L e =

1.5

L

L e =

1.0

L

L e =

0.8

5L

L e =

0.7

L

(b) (c) (d) (e)

σc,a,|| ≤ σc,adm,|| (6.27)

The applied compressive stress is calculated usingequation 6.24 and the permissible compressivestress is given by equations 6.25 or 6.26 dependingupon the slenderness ratio.

6.8.4.2 Members subject to axial compressionand bending

This category includes compression members sub-ject to eccentric loading which can be equated toan axial compression force and bending moment.According to clause 2.11.6 of BS 5268, memberswhich are restrained at both ends in position butnot direction, which covers most real situations,should be so proportioned that

σ

σ −σσ

+σ

σm,a,||

m,adm,||c,a,||

e12

c,a,||

c,adm,1

1.5 1

K

≤||

(6.28)

whereσm,a,|| applied bending stressσm,adm,|| permissible bending stressσc,a,|| applied compression stressσc,adm,|| permissible compression stress

(including K12)σe Euler critical stress = π2Emin/(Le/i )

2

Equation 6.28 is the normal interaction formula usedto ensure that lateral instability does not arise incompression members subject to axial force and

bending. Thus if the column was subject to com-pressive loading only, i.e. M = 0 and σm,a,|| = 0, thedesigner would simply have to ensure that σc,a,||/σc,adm,|| ≤ 1. Alternatively, if the column was subjectto bending only, i.e. F = σc,a,|| = 0, the designershould ensure that σm,a,||/σm,adm,|| ≤ 1. However, ifthe column was subject to combined bending andaxial compression, then the deflection as a result ofthe moment M would lead to additional bendingdue to the eccentricity of the force F as illustratedin Fig. 6.8. This is allowed for by the factor

1[1 (1.5 )/ ]c,a,|| 12 e− σ σK

in the above expression.

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301

SLENDERNESS RATIO

λ = Le/ i ⇒ L e = 1.0 × h = 1.0 × 3750 = 3750 mm

i

IA

dbdb

b = = = = =

3 2/1212

100

1228.867

λ =

375028.867

= 129.9 < 180 OK

GRADE STRESSES AND MODULUS OF ELASTICITYGrade GS redwood belongs to strength class C16 (Table 6.2). Values in N/mm2 are as follows

Compression parallel to grain Modulus of elasticityσc,g,|| Emin

6.8 5800

MODIFICATION FACTORK3, duration of loading is 1.0

Emin

c,σ ||

. .

.=×

=5800

6 8 1 0852 9 and λ = 129.9

From Table 6.6 by interpolation K12 is found to be 0.261.

Emin

σc, ||

λ

120 129.9 140

800 0.280 0.217852.9 0.293 0.261 0.228900 0.304 0.237

AXIAL LOAD CAPACITYPermissible compression stress parallel to grain is

σc,adm,|| = σc,g,||K3K12 = 6.8 × 1.0 × 0.261 = 1.77 N/mm2

Hence the long-term axial load capacity of column is

σc,adm,||A = 1.77 × 104 × 10−3 = 17.7 kN

Example 6.5 Timber column resisting an axial load (BS 5268)A timber column of redwood GS grade consists of a 100 mm square section which is restrained at both ends inposition but not in direction. Assuming that the actual height of the column is 3.75 m, calculate the maximum axiallong-term load that the column can support.


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302

Example 6.6 Timber column resisting an axial load and moment(BS 5268)

Check the adequacy of the column in Example 6.5 to resist a long-term axial load of 10 kN and a bending momentof 350 kN mm.

SLENDERNESS RATIOλ = L e/i = 129.9 < 180 (Example 6.5)

GRADE STRESSES AND MODULUS OF ELASTICITYValues in N/mm2 for timber of strength class C16 are as follows

Bending parallel Compression Modulus ofto grain parallel to grain elasticityσmg,|| σc,g,|| Emin

5.3 6.8 5800

MODIFICATION FACTORSK3 = 1.0

K

h7 100 .

. .

= = =300 300

1 1280 11 0 11

K12 = 0.261 (see Example 6.5)

COMPRESSION AND BENDING STRESSESPermissible compression stress is

σc,adm,|| = σc,g,||K3K12 = 6.8 × 1.0 × 0.261 = 1.77 N/mm2

Applied compression stress is

σc,a,|| =

axial loadA

=×10 10

10

3

4 = 1 N/mm2

Permissible bending stress is

σm,adm,|| = σm,g,||K3K7 = 5.3 × 1.0 × 1.128 = 5.98 N/mm2

Applied bending stress is

σm,a,|| =

MZ

=××

350 10167 10

3

3 = 2.10 N/mm2

Euler critical stress is

σ =

π=

π ×e

2min

e2

2

2

( / )

5800(129.9)

EL i

= 3.39 N/mm2

Since column is restrained at both ends, in position but not in direction, check that the column is so proportionedthat

σ

σ −σσ

+σ

σm,a,

m,adm,c,a,

e12

c,a,

c,adm,1

1.5 1||

||||

||

||K

≤

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preference, or on the loads they are required totransmit.

The frame is usually covered by a cladding ma-terial such as plasterboard which may be requiredfor aesthetic reasons, but will also provide lateralrestraint to the studs about the y–y axis. If the wallis not surfaced or only partially surfaced, the studsmay be braced along their lengths by internalnoggings. Bending about the x–x axis of the stud isassumed to be unaffected by the presence of thecladding material.

Since the centre-to-centre spacing of the studis normally less than 610 mm, the load-sharingfactor K8 will apply to the design of stud walls.The design of stud walling is illustrated in thefollowing example.

Fig. 6.9 Details of a typical stud wall: (a) elevation; (b) section; (c) typical fixing of top and bottom plates to studs.

6.9 Design of stud wallsIn timber frame housing the loadbearing walls arenormally constructed using stud walls (Fig. 6.9).These walls can be designed to resist not only thevertical loading but also loads normal to the walldue to wind, for example. Stud walls are normallydesigned in accordance with the requirements ofBS 5268: Part 6: Code of Practice for Timber FrameWalls; Section 6.1: Dwellings not exceeding four storeys.They basically consist of vertical timber members,commonly referred to as studs, which are held inposition by nailing them to timber rails or plates,located along the top and bottom of the studs. Themost common stud sizes are 100 × 50, 47, 38 mmand 75 × 50, 47, 38 mm. The studs are usuallyplaced at 400 or 600 mm centres depending upon

Studs cutsquareand tightlybutted toplates

yx

y

A minimum of two 90 mm longnails, offset to reduce risk of splitting

CladdingmaterialTop plate

Studs

Bottom (sole) plate

400 mm or 600 mm centre-to-centredistance between studs

(a) (b)

Noggings

x

(c)

Example 6.6 continuedSubstituting

2.10

5.98 1 1.5 1

3.39 0.261

1

1.77−×

×+

= 0.397 + 0.565 = 0.962 < 1

Therefore a 100 × 100 column is adequate to resist a long-term axial load of 10 kN and a bending moment of350 kN mm.

Design of stud walls

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SLENDERNESS RATIOEffective height

Lex = coefficient × L = 1.0 × 3750 = 3750 mm

Ley = coefficient × L /2 = 1.0 × 3750/2 = 1875 mm

Radius of gyration

i

IAxxxx

3

(1/12) 44 100

44 100

100

12= =

× ××

=

i

IAyyyy

3

(1/12) 100 44

44 100

44

12= =

× ××

=

Slenderness ratio

λ = =xx

ex

xx

Li

/

3750

100 12 = 129.9 < 180

λ = =yy

ey

yy

L

i

/

1875

44 12 = 147.6 < 180 (critical)

Note that where two values of λ are possible the larger value must always be used to find σc,adm,||.

Example 6.7 Analysis of a stud wall (BS 5268)A stud wall panel has an overall height of 3.75 m including top and bottom rails and vertical studs at 600 mm centreswith nogging pieces at mid-height. Assuming that the studs, rail framing and nogging pieces comprise 44 × 100 mmsection of strength class C22, calculate the maximum uniformly distributed long term total load the panel is able tosupport.

x 100 mmx

44 mmy

y600 mm 600 mm

1875

1875

3750

Noggings

Plasterboardcovering

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305

6.10 SummaryThis chapter has attempted to explain the con-cepts of stress grading and strength classes and theadvantages that they offer to designers and con-tractors alike involved in specifying timber for struc-tural purposes. The chapter has described the designof flexural and compression members and studwalling, to BS 5628: Part 2: Structural Use of Timber,

which is based on permissible stress principles. Inthe case of flexural members (e.g. beams, rafters andjoints), bending, shear and deflection are found tobe the critical factors in design. With compressionmembers (e.g. struts and columns), the slender-ness ratio has a major influence on load-carryingcapacity. Stud walls are normally designed on theassumption that the compression members act to-gether to support a common load.

Example 6.7 continuedGRADE STRESSES AND MODULUS OF ELASTICITYFor timber of strength class C22, values in N/mm2 are as follows:

Compression parallel to grain Modulus of elasticityσc,g,|| Emin

7.5 6500

MODIFICATION FACTORS

K3 = 1.0 K8 = 1.1

E EK

min

c,

min

c,g, 3

6500

7.5 1.0σ σ|| ||

= =×

= 866.7 and λ = 147.6

From Table 6.6 K12 = 0.212 by interpolation.

Emin

σc, ||

λ

140 147.6 160

800 0.217 0.172866.7 0.230 0.212 0.183900 0.237 0.188

AXIAL STRESSESPermissible compression stress parallel to grain σc,adm,|| is

σc,adm,|| = σc,g,||K3K8K12 = 7.5 × 1.0 × 1.1 × 0.212 = 1.75 N/mm2

Axial load capacity of stud is

σc,adm,||A = 1.75 × 44 × 100 × 10−3 = 7.7 kN

Hence uniformly distributed load capacity of stud wall panel is

7.7/0.6 = 12.8 kN/m

Note that the header spans 0.6 m and that this should also be checked as a beam in order to make sure that it iscapable of supporting the above load.

Summary

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306

Questions1. (a) Discuss the factors which influence

the strength of timber and explainhow the strength of timber is assessedin practice.

(b) A simply supported timber roofbeam spanning 5 m supports a totaluniformly distributed load of 11 kN.Determine a suitable section for thebeam using timber of strength classC16. Assume that the bearing lengthis 125 mm and that the compressionedge is held in position.

2. (a) Give typical applications of timber inthe construction industry and for eachcase discuss possible desirableproperties.

(b) Redesign the timber joists in Example6.2 using timber of strength class C22.

3. (a) Distinguish between softwood andhardwood and grade stress andpermissible stress.

(b) Calculate the maximum long termimposed load that a flat roof cansupport assuming the followingconstruction details:– roof joists are 50 mm × 225 mm

of strength class C16 at 600 mmcentres

– effective span is 4.2 m– unit weight of woodwool (50 mm

thick) is 0.3 kN/m2

– unit weight of boarding, bitumenand roofing felt is 0.45 kN/m2

– unit weight of plasterboard andskim is 0.22 kN/m2

4. (a) Discuss the factors accounted for bythe modification factor K12 in thedesign of timber compressionmembers.

(b) Design a timber column of effectivelength 2.8 m, capable of resisting thefollowing loading:(i) medium term axial load of

37.5 kN(ii) long term axial load of 30 kN

and a bending moment of300 kN mm.

5. (a) Explain with the aid of sketchesconnection details which will give riseto the following end conditions:(i) restrained in position and

direction(ii) restrained in position but not in

direction(iii) unrestrained in position and

direction.(b) Design a stud wall of length 4.2 m

and height 3.8 m, using timber ofstrength class C16 to support a long-term uniformly distributed load of14 kN/m.

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PART THREE

STRUCTURALDESIGN TO THEEUROCODES

Part Two of this book has described the design ofa number of structural elements in the four media:concrete, steel, masonry and timber to BS 8110,BS 5950, BS 5628 and BS 5268 respectively. Theprincipal aim of this part of the book is to describethe salient features of the structural Eurocodes forthese media, Eurocodes 2, 3, 6 and 5 respectively,and highlight the significant differences betweenthe British Standard and the correspondingEurocode.

The subject-matter has been divided into fivechapters as follows:

1. Chapter 7 introduces the Eurocodes and providesanswers to some general questions regarding theirnature, role, method of production and layout.

2. Chapter 8 describes the contents of Part 1.1 ofEurocode 2 for the design of concrete buildingsand illustrates the new procedures for designingbeams, slabs, pad foundations and columns.

3. Chapter 9 describes the contents of Part 1.1 ofEurocode 3 for the design of steel buildings andillustrates the new procedures for designingbeams, columns and connections.

4. Chapter 10 describes the contents of Part 1.1 ofEurocode 6 for the design of masonry structuresand illustrates the new procedures for the designof unreinforced load-bearing and panel walls.

5. Chapter 11 describes the contents of Part 1.1 ofEurocode 5 for the design of timber structuresand illustrates the new procedures for designingflexural and compression members.

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The structural Eurocodes: An introduction

308

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309

Chapter 7

The structuralEurocodes:An introduction

containing rules relevant to the design of a rangeof structures including buildings, bridges, waterretaining structures, silos and tanks. EN 1991 pro-vides characteristic values of loads (termed ‘actions’in Eurocode-speak) needed for design. EN 1990,the head Eurocode, is the world’s first material-independent design code and provides guidanceon determining the design value of actions andcombination of actions, including partial safety fac-tors for actions. EN 1997 covers the geotechnicalaspect of foundation design. EN 1998 is devotedto earthquake design and provides guidance onachieving earthquake resistance of buildings, bridges,towers, geotechnical structures, amongst others.

Eurocodes will have the same legal standing asthe national equivalent design standard or code ofpractice do eventually. Actually under UK law thereis no prescribed set of acceptable codes, merely alist of Approved Documents. Thus in due courseEurocodes will attain this status when it is expectedthat reference to BS codes will be omitted.

Eurocodes were published first as preliminarystandards, known as ENV (Norme VornormeEuropéenne), beginning in 1992. Now, after severalyears, following comments received on the prelim-inary versions, they have been revised and reissuedas full European standards, known as EN (NormeEuropéenne), together with a National Annex whichcontains supplementary information specific toeach Member State. Following a few years ofco-existence, the EN Eurocodes will become man-datory in the sense that all conflicting standardsmust be withdrawn. In this context ‘withdrawn’means that the codes will no longer be maintainedand in time will become obsolete.

7.2 Benefits of EurocodesThe establishment of international codes of prac-tice for structural design is not a new idea. Indeed,

This chapter provides a brief introduction to theEurocodes. It describes the nature, benefits and me-chanics of producing the Eurocodes. It describes thegeneral format of the documents and discusses somedifficulties associated with drafting these codes togetherwith how these difficulties were resolved. The roles ofthe National Annex and NCCIs in the Eurocode systemare also highlighted.

7.1 ScopeThe Eurocodes are a family of ten European codesof practice for the design of building and civilengineering structures in concrete, steel, timberand masonry, amongst other materials. Table 7.1lists the reference numbers and titles of the tenEurocodes. Like the present UK codes of practice,Eurocodes will come in a number of parts, each

Table 7.1 The structural Eurocodes

EuroNorm Titlereference

EN1990 Eurocode: Basis of designEN1991 Eurocode 1: Actions on structuresEN1992 Eurocode 2: Design of concrete

structuresEN1993 Eurocode 3: Design of steel structuresEN1994 Eurocode 4: Design of composite steel

and concrete structuresEN1995 Eurocode 5: Design of timber structuresEN1996 Eurocode 6: Design of masonry

structuresEN1997 Eurocode 7: Geotechnical designEN1998 Eurocode 8: Design of structures for

earthquake resistanceEN1999 Eurocode 9: Design of aluminium

structures

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310

the first draft of Eurocode 2 for concrete structureswas based on the CEB (Comité Européen duBeton) Model Code of 1978 produced by a num-ber of experts/enthusiasts from various Europeancountries. Similarly Eurocode 3 was based on the1977 Recommendations for Design of Steel Structurespublished by ECCS (European Convention forConstructional Steelwork), which was the workof several expert committees drawn from variouscountries from Europe and beyond. The originalmotivation for this work was to improve the artand science of structural design. However, the drivetowards the political and economic unification ofcountries in the EC has broadened the aims andobjectives of these documents, which no doubt hasalso influenced their final shape.

There are several advantages to be gained fromhaving design standards which are accepted by allMember States. The first and foremost reason isthat the provision of Eurocodes and the associatedEuropean standards for construction products willhelp lower trade barriers between the MemberStates. This will allow contractors and consultantsfrom all Member States to compete fairly for workwithin Europe as well as help facilitate the market-ing and use of structural materials, components,kits and design aids/software, right across Europe.Hopefully this will lead to a pooling of resourcesand the sharing of expertise, thereby loweringproduction costs. It is further believed that theirexistence will boost the international standing ofEuropean engineers which should help in increasingtheir chances of winning work abroad. A furtherbenefit of having Eurocodes is that they will makeit easier for European engineers to practise through-out Europe.

7.3 Production of EurocodesGenerally, each Eurocode has been drafted by asmall team of experts from various Member States.These groups were formerly under contract to theEC Commission but in 1989 responsibility forpreparation and publication of the Eurocodes passedto CEN (Comité Européen de Normalisation), theEuropean Standards Organisation, whose membersare the National Standards Bodies, e.g. BritishStandards Institute in the UK. During the ENVstage of production a liaison engineer from eachMember State was involved in evaluating the finaldocument and discussing with the drafting teamthe acceptability of the Eurocode in relation to thenational code. During the conversion to EN status,

BSI has established mirror committees for eachEurocode, through which all UK comments arerouted. These committees consider both the tech-nical aspects as well as the economic impacts of theEurocodes and manage the task of calibrating thecode.

7.4 FormatFig. 7.1 shows the general layout of each Eurocode.Each consists of the full text of the Eurocode,including any annexes approved by CEN, whichis normally preceded by a national title page andnational foreword. The latter lists the nationalstandard(s) having the same scope as the Eurocodeand provides some information on the accompany-ing National Annex, e.g. scope, publisher, generaldate of availability.

7.5 Problems associated withdrafting the Eurocodes

Inevitably, there were many problems faced bycommittee members responsible for drafting theEurocodes. These centred on agreeing a commonterminology acceptable to all the Member States,resolving differing opinions on technical issues,taking into account national differences in con-struction materials and products and design andconstruction practices, and regional differences inclimatic conditions. In addition, it was also con-sidered essential that all Eurocodes should be com-prehensive yet concise. The following subsectionsoutline how some of these issues were resolved with-out compromising the clarity or, indeed, simplicityof the codes.

7.5.1 TERMINOLOGYAt the outset of this work it was necessary tostandardise the terminology used in the Euro-codes. Generally, this is similar to that found inthe equivalent UK national standards. However,there are some minor differences; for example, aspreviously noted, loads are now called actions whiledead and live loads are referred to as permanentand variable actions, respectively. Similarly, bend-ing moments and axial loads are now called inter-nal moments and internal forces respectively. Thesechanges were adopted to avoid problems duringtranslation of the codes. They are so minor thatthey are unlikely to present any problems to UKengineers.

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311

Problems associated with drafting the Eurocodes

A: National title pageB: National Foreword

C: CEN title pageD: Main textE: Annexes

A

D

B

C

D

D

E

BRITISH STANDARD BS EN 1993-1-1

Eurocode 3: Design ofsteel structures –

Part 1-1: General rules andrules for Buildings

BSi

Fig. 7.1 Layout of Eurocodes

7.5.2 PRINCIPLES AND APPLICATION RULESIn order to produce documents which are (a)concise, (b) describe the overall aims of designand (c) provide specific guidance as to how theseaims can be achieved in practice, the materialin the Eurocodes is divided into ‘Principles’ and‘Application rules’.

Principles comprise general statements, defini-tions, requirements and models for which no alter-native is permitted. Principles are indicated by theletter P after the clause number. The Applicationrules are generally recognised rules which followthe statements and satisfy the requirements given inthe Principles (Fig. 7.2). The absence of the letterP after the clause number indicates an Applicationrule. The use of alternative application rules to thoserecommended in the Eurocode is permitted providedit can be shown that the alternatives are at leastequivalent and do not adversely affect other designrequirements. It is worth noting, however, that ifan alternative Application rule is used the resultingdesign will not be deemed Eurocode compliant.

7.5.3 NATIONALLY DETERMINED PARAMETERSAND NATIONAL ANNEXES

Possible differences in construction material/products and design and construction practices, and

Fig. 7.2 Example of Principle and Application Rule(Cl 2.4 Durability, EN 1990, pp. 26–27)

7.3.2 Minimum reinforcement areas

(1)P If crack control is required, a minimum amount ofbonded reinforcement is required to control cracking inareas where tension is expected. The amount may beestimated from equilibrium between the tensile force inconcrete just before cracking and the tensile force inreinforcement at yielding or at a lower stress if necessaryto limit the crack width.

(2) Unless a more rigorous calculation shows lesser areas tobe adequate, the required minimum areas of reinforcementmay be calculated as follows. In profiled cross sections likeT-beams and box girders, minimum reinforcement should bedetermined for the individual parts of the section (webs,flanges).

As,minσs = k ckfct,effAct (7.1)

regional differences in climatic conditions, e.g. windand snow loading, has meant that some parameters,e.g. partial safety factors, allowance in design fordeviation of concrete cover, a particular methodor application rule if several are proposed in theEN, etc., may be determined at the national level.Where a national choice is allowed this is indicated

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in a note in the normative text under the relevantclause. The note may include the recommendedvalue of the parameter, or preferred method, etc.but the actual value, methodology, etc., to be usedin a particular Member State will be given in anaccompanying National Annex. In the UK, BSI isresponsible for producing and publishing NationalAnnexes. The recommended values of these para-meters and design method/procedures are collectivelyreferred to as Nationally Determined Parameters(NDPs). The NDPs determine various aspects ofdesign but perhaps most importantly the level ofsafety of structures during execution (Eurocodeterminology for construction/fabrication) andin-service, which remains the responsibility of indi-vidual nations. The European Commission wantsthe NDPs to be harmonised so that their numberis kept to a minimum.

7.5.4 ANNEXESSome procedures which are not used in everydaydesign have been included in annexes. Some ofthe annexes are labelled ‘normative’ while othersare labelled ‘informative’. The majority fall into thelatter category. The material which appears in the‘normative’ annexes has the same status as the restof the Eurocode but appears there rather than thebody of the code in order to make the documentas ‘user friendly’ as possible. The material in the‘informative’ annexes, however, does not have anystatus but has been included merely for informa-tion. The National Annex should be consulted fordecisions on the use of informative annexes in acountry.

7.5.5 NON CONTRADICTORY COMPLEMENTARYINFORMATION (NCCI)

To assist designers to use Eurocodes, the systemallows the National Annex to refer to sourcesof non-contradictory complementary information(NCCI). An example of an NCCI relevant to con-crete design is Published Document (PD) 6687. Itwas prepared and published by BSI and includesthe following:

• background to the values chosen for some ofthe NDPs in EN 1992

• commentary on some specific code clauses• requirements, which are additional to those in

the code and National Annex, to comply withthe UK Building Regulations.

A number of NCCIs for steel design have beenproduced by the Steel Construction Institute andwill be referred to in Chapter 9 of this book.

7.6 Decimal pointIn the Eurocodes the decimal point has beenreplaced by a comma. However, for consistencywith the other chapters in this book the use of thedecimal point has been retained.

7.7 ImplementationAll the Eurocode Parts have now been publishedbut guidance on the timing and the circumstancesunder which the Eurocodes must be used is stillnot available. Present indications would suggest thatEurocodes will be listed in the Approved Docu-ments as a means of meeting the requirements ofthe Building Regulations by 2010. Some clientorganisations, e.g. the UK Highways Agency andthe Rail Standards Authority, are keen to incor-porate Eurocodes in their specifications as soon aspossible. Nevertheless, if the experiences of thosefamiliar with similar changes in design practice isanything to go by this transition will occur over amuch longer time frame. For example it took atleast ten years to introduce CP110 into the designoffice. Hearsay evidence suggests that some engi-neers still continue to submit calculations to BS 449for building approval despite the publication ofBS 5950 in 1985.

7.8 MaintenanceThe precise mechanism by which the Eurocodeswill be maintained remains unclear. All that can besaid on this point is that under CEN rules all ENsshould be reviewed every five years.

7.9 Difference between nationalstandards and Eurocodes

Inevitably there are many differences betweenEurocodes and the national codes. This is boundto cause confusion and will no doubt increase designtime. The change in technical approval proceduresare fairly minor, thanks largely to the work put inby the UK members of the various drafting panels.Consequently it should not take much time forengineers to become familiar with design pro-cedures in the Eurocodes and subsequently to usethem. The problem is not so much about usingthe design rules but finding them in the firstplace. Designers will now have to refer to a muchlarger number of documents in order to design the

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Difference between national standards and Eurocodes

simplest of structures. For example to design asteel beam a designer will now have to refer to EN1990 to determine the value of design loading, EN1993-1-1 for design guidance on bending and shear,and EN 1993-1-5 for web and stiffener design.Connection design has been separated into EN1993-1-8. Currently all the design rules are avail-able in one document: BS 5950: Part 1. Moreoverthe design rules in Eurocodes are sequenced onthe basis of structural phenomena rather than typeof element, which means that designers must have

a good understanding of structural behaviour inorder to use the code.

Having discussed these more general aspects,the following chapters will describe the contents ofEN 1992, EN 1993, EN 1995 and EN 1996 for,respectively, concrete, steel, timber and masonrydesign, and the significant differences betweenthe Eurocode and the corresponding British Stand-ard. Chapter 8 on concrete design to EN 1992also includes a discussion on EN 1990 and EN1991.

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Eurocodes 2: Design of concrete structures

314

design values of actions (section 8.5), BS 4449 formechanical properties of reinforcing steel (section8.4.1), Part 1.2 of Eurocode 2 for fire design(section 8.7.1), BS 8500 and EN 206 for durabilitydesign (section 8.7.2) and Eurocode 7 for founda-tion design (Fig. 8.1). The main reason cited forstructuring the information in this way is to avoidrepetition and make the design guidance in Part1.1 more concise than BS 8110.

Part 1.1 of Eurocode 2, hereafter referred to asEC 2, was issued as a preliminary standard or ENVin 1992 and in final form as BS EN 1992-1-1 in2004. The following subjects are covered in EC 2:

Section 1: GeneralSection 2: Basis of designSection 3: MaterialsSection 4: Durability and cover to reinforcementSection 5: Structural analysisSection 6: Ultimate limit statesSection 7: Serviceability limit statesSection 8: Detailing of reinforcement and

prestressing tendons – GeneralSection 9: Detailing of members and particular

rulesSection 10: Additional rules for precast concrete

elements and structuresSection 11: Lightweight aggregate concrete

structuresSection 12: Plain and lightly reinforced concrete

structures

Also included are ten annexes which provide supple-mentary information on a range of topics includingcreep and shrinkage, reinforcing steel, durabilitydesign and analysis of flat slabs and shear walls.

The purpose of this chapter is to describe thecontents of EC 2 and to highlight the principaldifferences between EC 2 and BS 8110. A numberof examples covering the design of beams, slabs, padfoundation and columns have also been includedto illustrate the new design procedures.

This chapter describes the contents of Part 1.1 ofEurocode 2, the new European standard for the designof buildings in concrete, which is expected to replaceBS 8110 by about 2010. The chapter highlights theprincipal differences between Eurocode 2: Part 1.1 andBS 8110 and illustrates the new design procedures bymeans of a number of worked examples on beams, slabs,pad foundation and columns. To help comparison, butprimarily to ease understanding of the Eurocode, thematerial has been presented in a way familiar to usersof BS 8110, rather than strictly adhering to the sequenceof chapters and clauses adopted in the Eurocode.

8.1 IntroductionEurocode 2 applies to the design of buildings andcivil engineering works in concrete. It is based onlimit state principles and comes in four parts asshown in Table 8.1.

Part 1.1 of Eurocode 2 gives a general basis forthe design of structures in plain, reinforced, light-weight, pre-cast and prestressed concrete. In addi-tion, it gives some detailing rules which are mainlyapplicable to ordinary buildings. It is largely sim-ilar in scope to BS 8110 which it will replace byabout 2010. Design of building structures cannotwholly be undertaken using Part 1.1 of Eurocode2, however. Reference will have to be made toa number of other documents, notably EN 1990(Eurocode 0) and Eurocode 1 to determine the

Table 8.1 Scope of Eurocode 2: Design ofConcrete Structures

Part Subject

1.1 General rules and rules for buildings1.2 Structural fire design2 Reinforced and prestressed concrete bridges3 Liquid retaining and containment structures

Chapter 8

Eurocodes 2:Design of concretestructures

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8.2 Structure of EC 2Although the ultimate aim of EC 2 and BS 8110is largely the same, namely to principally provideguidance on the structural use of reinforced andprestressed concrete in building structures, theorganisation of material in the two documents israther different.

For example, BS 8110 contains separate chapterson the design of beams, slabs, columns, bases, etc.However, EC 2 divides the material on the basisof structural action, i.e. bending, shear, deflection,torsion, etc., which may apply to any element.Furthermore, prestressed concrete is not dealt withseparately in EC 2 as in BS 8110, but each sectionon bending, shear, deflection, etc., contains rulesrelevant to the design of prestressed members.

There is a slight departure from this principle inchapters 5 and 9 of EC 2 which give guidance onthe analysis and detailing respectively of specificmember types.

8.3 SymbolsThe following symbols which have largely beentaken from EC 2 have been used in this chapter.

GEOMETRIC PROPERTIES:b width of sectiond effective depth of the tension

reinforcementh overall depth of sectionx depth to neutral axis

Fig. 8.1 Eurocode 2: Part 1.1 supporting documents.

z lever armleff effective span of beams and slabsln clear distance between the faces of the

supportsa1, a2 allowance at supports used for

calculating the effective span of amember

cnom nominal cover to reinforcementd2 effective depth to compression

reinforcement

BENDING:gk, Gk characteristic permanent actionqk, Qk characteristic variable actionwk, Wk characteristic wind loadz lever armfck characteristic compressive cylinder

strength of concrete at 28 daysfyk characteristic yield strength of

reinforcementfcd design compressive strength of concretefyd design yield strength of reinforcementFd design actionFk characteristic actionFrep representative actionXk characteristic strengthXd design strengthγC partial factor for concreteγ f, γF partial factor for actionsγS partial factor for steelγG partial factor for permanent actionsγQ partial factor for variable actionsγM partial factor for material properties

Symbols

Geotechnicaldesign

Properties ofreinforcing steel

Durabilitydesign

Firedesign

Actions

Eurocode 2:Part 1.1

BS 8500

EN 206

Eurocode 2:Part 1.2

Eurocode 1:Part 1

Eurocode 0

BS 4449

Eurocode 7

Buildingstructure

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αcc a coefficient taking account of sustainedcompression

C concrete strength classK0 coefficient given by M/fckbd2

K0′ coefficient given by MRd/fckbd2 = 0.167MEd design ultimate momentMRd design ultimate moment of resistanceAs1 area of tension reinforcementAs2 area of compression reinforcement

SHEAR:k a coefficient relating to section depths spacing of stirrupsz ≈ 0.9dα angle of shear reinforcement to the

longitudinal axisθ angle of the compression strut to the

longitudinal axisρ1 reinforcement ratio corresponding to As1

ρw,min minimum shear reinforcement ratioν1 strength reduction factor for concrete

cracked in shearσcp average compressive stress in concrete

due to axial forcefywd design yield strength of shear

reinforcementVEd design value of the applied shear forceVRd,c shear resistance of member without

shear reinforcementVRd,max maximum shear force, limited by

crushing resistance of compressionstrut

VRd,s shear force sustained by yielding ofshear reinforcement

Asw cross-sectional area of the shearreinforcement

COMPRESSION:b width of cross-sectionh depth of cross-sectionl clear height between end restraintslo effective lengthk1, k2 relative flexibilities of rotational

restraints at column endsi radius of gyrationλlim limiting slenderness ratio for ignoring

second order effectsθ inclination used to represent

imperfections of the structureeo minimum eccentricity = h/30 but not

less than 20 mmM01,M02 respectively the numerically lower and

higher values of the first order end

moments at ultimate limit states, i.e.(|M01| ≤ |M02|) including allowance forimperfections

ei eccentricity due to geometricalimperfections

e2 deflection due to second order effectsNEd design axial loadn = NEd/Ac fcd

nu = 1 + ω where ω = As fyd/(Ac fcd)nbal = 0.4M0Ed equivalent first order moment including

the effect of imperfections (= M0e)MEd design bending momentAc cross-sectional area of the concreteAs area of longitudinal reinforcement

8.4 Material properties

8.4.1 CHARACTERISTIC STRENGTHS

Concrete (Cl. 3.1, EC 2)Unlike BS 8110, the design rules in EC 2 arebased on the characteristic (5 per cent) compressivecylinder strength of concrete at 28 days ( fck).Equivalent cube strengths ( fck,cube) are included inEC 2 but they are only regarded as an alternativemethod to prove compliance. Generally, the cylinderstrength is approximately 0.8 × the cube strengthof concrete i.e. fck ≈ 0.8 × fck,cube. The design rulesin EC 2 are valid for concrete of strength classesup to C90/105, i.e. cylinder strength 90 MPa andcube strength 105 MPa. However, this book onlydiscusses the rules relevant to concrete strengthclasses up to C50/60 as these are the classes mostoften used in reinforced concrete design.

Table 8.2 shows the actual strength classes com-monly used in reinforced concrete design. Alsoincluded in the table are two formulae for calculat-ing the mean tensile strength ( fctm) and the 5 percent fractile tensile strength ( fctk,0.05) of concrete,which are referred to later in this chapter.

Reinforcing steel (Cl. 3.2, EC 2)According to Annex C, the design rules in EC 2are applicable to steel reinforcement with character-istic yield strength in the range 400–600 N mm−2.Details of the actual yield strength of steel availablein the UK for the reinforcement of concrete can befound in BS 4449: 2005. This document indicatesthat steel reinforcement will now be manufacturedin three grades, all of 500 N mm−2 characteristicyield strength, but with differing ductility (Table8.3). Plain round bars of characteristic yield strength

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Table 8.2 Concrete strength classes and properties (based Table 3.1, EC 2)

Strength Classes for Concrete C20/25 C25/30 C28/35 C30/37 C32/40 C35/45 C40/50 C45/55 C50/60

fck 20 25 28 30 32 35 40 45 50fck,cube 25 30 35 37 40 45 50 55 60

fctm fctm = 0.3 × f ck(2/3) N mm−2 (8.1)

fctk,0.05 fctk,0.05 = 0.7 × fctm N mm−2 (8.2)

250 N mm−2 are not covered in this standard, andthis will presumably cease to be produced in theUK. Present indication would suggest that ductilityclasses B and C will be the most widely availableand specified steel in the UK. Ductility class A, insizes 12 mm and below, in coil form is widely usedby reinforcement fabricators for use on automaticlink bending machines (see CARES informationsheet on ‘Design, manufacture and supply of rein-forcement steel’).

8.4.2 DESIGN STRENGTHS (CL. 2.4.2.4, EC 2)Generally, design strengths (Xd) are obtained bydividing the characteristic strength (Xk) by theappropriate partial safety factor for materials (γM):

X

Xd

k

M

= γ

(8.3)

The partial safety factors for concrete and steelreinforcement for persistent and transient designsituations are shown in Table 8.4. Note that in EC2 the partial safety factor for ultimate limit stateshas a single value of 1.5. This is different to BS8110 where the partial safety factor is a function ofthe stress type under consideration (see Table 3.3,Chapter 3).

8.5 Actions (Cl. 2.3.1, EC 2)Action is the Eurocode terminology for load. EC 2defines an action (F) as a set of forces, deformations(e.g. differential settlement and temperature effects)or accelerations acting on the structure. Actionsmay be ‘permanent’ (G), e.g. self-weight of struc-ture, fittings and fixed equipment, or ‘variable’ (Q),e.g. weight of occupants, wind and snow loads.The following discusses how the characteristicand design values of actions on building structuresnot involving geotechnical actions for the ultimatelimit state of equilibrium and strength are assessed.Actions on/from structures involving geotechnicalactions, e.g. retaining walls and foundations areassessed somewhat differently and the reader isreferred to EN 1990 and Eurocode 7 for furtherdetails.

8.5.1 CHARACTERISTIC ACTIONS(CL. 2.3.1.1, EC 2)

The characteristic values of both permanent andvariable actions (Fk) are specified in BS EN 1991:Actions on Structures (EC 1). EC 1 includes materialtraditionally found in documents such as BS 6399:Loading for buildings and BS 648: Schedule for weightsof building materials, but its scope is far wider asit provides information on actions relevant to thedesign of all civil engineering structures, not justbuildings. EC 1 is published in ten parts as listedin Table 8.5. Parts 1-1 to 1-7 are the most relevantto the design of building structures.

Table 8.4 Partial safety factors for materials(based on Table 2.1N, EC 2)

Limit state γC for concrete γS for reinforcing andprestressing steel

Ultimate 1.5 1.15Serviceability 1.0 1.0

Table 8.3 Tensile and other properties of steelfor the reinforcement of concrete

Property Ductility Class

A B C

Characteristic yield strength, 500fyk (N mm−2)Young’s modulus (kN mm−2) 200Characteristic strain at ≥ 2.5 ≥ 5.0 ≥ 7.5ultimate force, εuk (%)

Actions

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8.5.2 DESIGN VALUES OF ACTIONS(CL. 2.4.1, EC 2 / CL. 6, EN 1990)

In general, the design value of an action (Fd) isobtained by multiplying the representative value(Frep) by the appropriate partial safety factor foractions (γf):

Fd = γf.Frep (8.4)

Tables 8.6 and 8.7 show the recommendedvalues of partial safety factor for permanent, γG,and variable actions, γQ, for the ultimate limit statesof equilibrium (EQU) and strength (STR). It canbe seen that the maximum values of γG and γQ are1.35 and 1.5 respectively. The comparable valuesin BS 8110 are 1.4 and 1.6. It can also be seenthat the partial safety factors for actions depend ona number of other aspects including the categoryof limit state as well as the effect of the actionon the design situation under consideration. Forexample, when checking for the limit states of

Table 8.5 Scope of Eurocode 1: Actions onStructures

Document No Subject

BS EN 1991-1-1 Densities, self-weight and imposedloads

BS EN 1991-1-2 Actions on structures exposed tofire

BS EN 1991-1-3 Snow loadsBS EN 1991-1-4 Wind loadsBS EN 1991-1-5 Thermal actionsBS EN 1991-1-6 Actions during executionBS EN 1991-1-7 Accidental actions due to impact

and explosionsBS EN 1991-2 Traffic loads on bridgesBS EN 1991-3 Actions induced by cranes and

machineryBS EN 1991-4 Actions in silos and tanks

Table 8.7 Load combinations and partial safety/combination factors for the ultimate limit state ofstrength

Limit state/Load combination Load Type

Permanent, Gk Imposed, Qk Wind, Wk

Unfavourable Favourable Unfavourable Favourable

Strength:1. Permanent and variable 1.35/1.35ξ* 1.0 1.5 0 –2. Permanent and wind 1.35/1.35ξ 1.0 – – 1.53. Permanent, imposed and wind

(a) 1.35 1.0 1.5ψ0,1 1.5ψ0,2

(b) 1.35/1.35ξ 1.0 1.5 0 1.5ψ0

(c) 1.35/1.35ξ 1.0 1.5ψ0 0 1.5

* See equation (8.9) later

Table 8.6 Partial safety factors for the ultimate limit state of equilibrium

Limit state Load Type


Unfavourable Favourable Unfavourable Favourable Unfavourable

Equilibrium 1.10 0.9 1.5 0 1.5Ψ0

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equilibrium and strength, the maximum values ofγG are 1.1 and 1.35, respectively. However, whenchecking for equilibrium alone, γG is taken to be1.1 if the action increases the risk of instability(unfavourable action) or 0.9 if the action reducesthe risk of instability (favourable action). For agiven limit state several combinations of loadingmay have to be considered in order to arrive at thevalue of the design action on the structure (seeTable 8.7) and guidance on this aspect of design isdiscussed in section 8.5.3.

In equation 8.4, Frep may be the characteristicvalue of a permanent or leading variable action(Fk), or the accompanying value (ΨFk) of a variableaction. In turn, the accompanying value of a vari-able action may be the combination value (Ψ0Fk),the frequent value (Ψ1Fk) or the quasi-permanentvalue (Ψ2Fk). The frequent value and the quasi-permanent values are used to determine valuesof accidental actions, e.g. impact and explosions,and to check serviceability criteria (deflection andcracking). The combination value is given by

Combination value = Ψ0Fk (8.5)

where Ψ0 is the combination factor obtained fromTable 8.8 and is a function of the type of variableaction. The factor Ψ0 has been introduced to takeaccount of the fact that where a structure is sub-ject to, say, two independent variable actions, it isunlikely that both will reach their maximum valuesimultaneously. Under these circumstances, it isassumed that the ‘leading’ variable action (i.e. Qk,1)is at its maximum value and any ‘accompanying’variable actions will attain a reduced value, i.e.Ψ0Qk,i, where i > 1. Leading and accompanyingvariable actions are assigned by trial and error asdiscussed below.

8.5.3 COMBINATION EXPRESSIONSHaving determined the design values of individualactions acting on the structure it is necessary to con-sider the effect of possible combination of actions.According to EN 1990 the design value of actioneffects, Ed, assuming the structure is subjected toboth permanent and a single variable action (e.g.dead load plus imposed load or dead load pluswind load) can be assessed using the followingexpression

Ed =

γ G,j k,jG

j ≥∑

1

“+” γQ,1Qk,1 (8.6)

where“+” implies ‘to be combined with’∑ implies ‘the combined effect of ’

Using the partial safety factors given in Table 8.7,the design value of the action effect is given by

Ed = 1.35Gk + 1.5Qk

(load combinations 1 and 2, Table 8.7)

The design value of an action effect due to per-manent and two (or more) variable actions, e.g.dead plus imposed and wind load, is obtained fromequation 8.7.

Ed =

γ G,j k,jGj ≥∑

1

“+” γQ,1Qk,1

“+”

γ ψQ,i i k,i01

, Qi f∑ (8.7)

Note that this expression yields two (or more)estimates of design actions and the most onerousshould be selected for design. For example, if astructure is subjected to permanent, office and windloads of Gk, Qk and Wk the values of the designactions are:

Ed = 1.35Gk,j “+” 1.5Q k “+” 1.5 × 0.5Wk

(load combination 3(b), Table 8.7 )

and

Ed = 1.35Gk,j “+” 1.5 × 0.7Q k “+” 1.5Wk

(load combination 3(c), Table 8.7)

Equations 8.6 and 8.7 are based on expression6.10 in EN 1990. This document also includestwo alternative expressions, namely 6.10a and 6.10b(reproduced as equations 8.8 and 8.9 respectively)for calculating the design values of actions, use ofwhich may improve structural efficiency, particularlyfor heavier structural materials such as concrete.

Table 8.8 Ψ0, Ψ2 values

Variable actions Ψ0 Ψ2

Imposed loadsDwellings 0.7 0.3Offices 0.7 0.3Shopping and congregation areas 0.7 0.6Storage 1.0 0.8Parking (vehicle weight ≤ 30 kN) 0.7 0.6

Snow loads (where latitudes 0.5 0.0≤ 1000 m above sea level)

Wind loads 0.5 0.0

Actions

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Fig. 8.2 Load Set 2: (a) all spans carrying the design permanent and variable loads and (b) alternate spans carrying thedesign permanent and variable load, other spans carrying only the design permanent load.

Ed =

γ G,j k,jGj≥∑

1

“+” γQ,1ψ0,1Qk,1

“+”

γ ψQ,i i k,i01

, Qi f∑ (8.8)

Ed =

ξj G,j k,jγ Gj≥∑

1

“+” γQ,1Qk,1

“+”

γ ψQ,i i k,i01

, Qi f∑ (8.9)

whereξ is a reduction factor for unfavourable

permanent actions. The value of ξrecommended in the National Annexto EC 2 is 0.925.

Note that equation 8.8 yields only one estimateof Ed (i.e. load combination 3(a) in Table 8.7)whereas equation 8.9 yields two (i.e. load combina-tions 3(b) and 3(c) in Table 8.7). For UK buildingstructures, designers may use the output from eitherequation 8.6 or 8.7 (depending on the numberof variable actions present) or the more onerousoutput from equations 8.8 and 8.9.

Use of actions determined via equations 8.6 / 8.7should lead to designs with comparable levels ofsafety to that currently achieved using BS 8110. How-ever, use of equations 8.8 and 8.9 may improvestructural efficiency, as illustrated in example 8.1.

8.5.3 LOADING ARRANGEMENTSAs discussed in section 3.6.2, design of continuousbeams (and slabs) will require consideration of load

arrangements. This term refers to the way in whichvariable actions should be positioned in order toproduce the most onerous design conditions. TheNational Annex to EC 2 suggests two alternativeloading arrangements for continuous beams: LoadSet 1 and Load Set 2. Generally, Load Set 2 (Fig.8.2) will be used in the UK as it is similar toexisting recommendations in BS 8110 (i.e. all spansfully loaded and maximum and minimum designloads on alternate spans) and requires considera-tion of a fewer number of load cases. However,unlike BS 8110, clause 2.4.3 of EC 2 recommendsthat the same value of γG (either 1.0 or 1.35 which-ever gives the most unfavourable effect) is used forall spans.

Load Set 2(a) can also be used to determine thedesign conditions on continuous slabs provided thefollowing assumptions are satisfied:

(i) In a one-way spanning slab the area of eachbay exceeds 30 m2 (see Fig. 3.62 for definitionof bay)

(ii) The ratio of the characteristic variable loadto the characteristic permanent load does notexceed 1.25

(iii) The characteristic imposed load does notexceed 5 kNm−2, excluding partitions.

According to the National Annex to EC 2 whenslabs are analysed using this load arrangement, theresulting support moments except those at the sup-ports of cantilevers should be reduced by 20 percent with a consequential increase in the spanmoments (Table 1 of National Annex to EC 2).

(a)

(b)

γGGk + γQQk

γGGkγGGk + γQQk

L L L

L L L

γGGkγGGk + γQQk

L L L

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Example 8.1 Design actions for simply supported beam (EN 1990)A simply supported beam for an office building has a span of 6 m. Calculate the values of the design bendingmoments, ME,d, assuming

(a) the beam supports uniformly distributed permanent and variable actions of 5 kNm−1 and 6 kNm−1 respectively(b) in addition to the actions described in (a) the beam also supports an independent variable concentrated load of

20 kN at mid-span.

LOAD CASE A

Actions

Since the beam is subjected to only one variable action use equation 8.6 to determine Ed where

Ed =

γ G,j k, jGj≥∑

1

“+” γQ,1Q k ,1

⇒ FE,d = 1.35 × (5 × 6) + 1.5 × (6 × 6) = 94.5 kN

Hence, ME,d =

F LE,d kNm8

94 5 6

870 9

. . =

×=

An alternative estimate of ME,d can be obtained using equations 8.8 and 8.9, respectively

Ed =

γ G,j k, jGj≥∑

1

“+” γQ,1ψ0,1Q k ,1 “+”

γ ψQ,i i k,i01

, Qif∑

⇒ FE,d = 1.35 × (5 × 6) + 1.5 × 0.7 × (6 × 6) + 0 = 78.3 kN

Ed =

ξ γj G, j k, jGj≥∑

1

“+” γQ ,1Qk ,1 “+”

γ ψQ,i i k,i01

, Qif∑

⇒ FE,d = 0.925 × 1.35 × (5 × 6) + 1.5 × (6 × 6) + 0 = 91.5 kN (critical)

Hence FE,d is 91.5 kN and ME,d =

F LE,d kNm.8

91 5 6

868 6

. . =

×=

LOAD CASE B

gk = 5 kN/m; qk = 6 kN/m

L = 6 m

Qk = 20 kNgk = 5 kN/m; qk = 6 kN/m

L = 6 m

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322

The extra complication here is that it is not clear if qk or Q k is the leading variable action. This can only be determinedby trial and error. This time use equation 8.7 to evaluate Ed, since there are two independent variable actions arepresent.Assuming qk is the leading variable action gives

Ed =

γ G,j k, jGj≥∑

1

“+” γQ,1Qk ,1 “+”

γ ψQ,i i k,i01

, Qif∑

FE,d = [1.35 × (5 × 6) + 1.5 × (6 × 6)] + 1.5 × 0.7 × 20

= 94.5(FE,d1) + 21(FE,d2) = 115.5 kN

and ME,d =

F L F LE,d E,d2 kNm1

8 4

94 5 6

8

21 6

4102 4

.

. + =

×+

×=

Assuming Q k is the leading variable action gives

FE,d = 1.35 × (5 × 6) + 1.5 × 20 + 1.5 × 0.7 × (6 × 6)

= 40.5(FE,d1) + 30(FE,d2) + 37.8(FE,d3) = 108.3 kN

and ME,d =

( )

( . . )

.

F F L F LE,d1 E,d3 E,d2 kNm+

+ =+ ×

+×

=8 4

40 5 37 8 6

8

30 6

4103 7 (maximum moments)

Alternatively use equations 8.8 and 8.9 to estimate FE,d. Assuming qk is the leading variable action and substitutinginto 8.8 gives

Ed =

γ G,j k, jGj≥∑

1

“+” γQ,1ψ0,1Q k ,1 “+”

γ ψQ,i i k,i01

, Qif∑

FE,d = 1.35 × (5 × 6) + 1.5 × 0.7 × (6 × 6) + 1.5 × 0.7 × 20 = 99.3 kN

FE,d is unchanged if Q k is taken as the leading variable action and in both cases ME,d = 90.2 kNm.Repeating this procedure using equation 8.9 and assuming, first, that qk is the leading variable action and Q k is theaccompanying variable action and, second, Q k is the leading variable action and qk is the accompanying variableaction gives, respectively

Ed =

ξ j G, j k, jγ Gj≥∑

1

“+” γQ,1Q k ,1 “+”

γ ψQ,i i k,i01

, Qif∑

FE,d = [0.925 × 1.35 × (5 × 6) + 1.5 × (6 × 6)] + [1.5 × 0.7 × 20]

= 91.5 + 21 = 112.5 kN

and ME,d =

F L F LE,d E,d kNm1 2

8 4

91 5 6

8

21 6

4100 1

.

. + =

×+

×=

FE,d = [0.925 × 1.35 × (5 × 6) + 1.5 × 20 + 1.5 × 0.7 × (6 × 6)

= 37.5 + 30 + 37.8 = 105.3 kN

and ME,d =

( )

( . . )

.

F F L F LE,d1 E,d3 E,d2 kNm+

+ =+ ×

+×

=8 4

37 5 37 8 6

8

30 6

4101 5 (maximum moment)

Again, the most structurally economical solution is found via equation 8.9, which will normally be the case forconcrete structures provided that permanent actions are not greater than 4.5 times variable actions except forstorage loads. However, this saving has to be weighed against the additional design effort required. Moreover, theoutput from equations 8.8 and 8.9 should not be used to perform stability calculations and the reader is referred toEN1990 for further information on this aspect. Note that the value of 1.35 for γG is conservative and used throughoutthis chapter.


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Fig. 8.3 Parabolic-rectangular stress-strain diagram for concrete ( fck ≤ 50 N mm−2) in compression (Fig. 3.3, EC 2).

Stress-strain diagrams

εc2

εc

εc2

σc

σc = fcd [1 – (1 –n

for 0 ≤ εc ≤ εc2

εcu2

Par

abol

ic a

xis

fck

fcd

Idealiseddesign

Idealisedcharacteristic

εc

8.6 Stress-strain diagrams

8.6.1 CONCRETE (CL. 3.1.7, EC 2)Figure 8.3 shows the idealised (characteristic anddesign) stress-strain relationships for concrete incompression. The basic shapes of the curves, i.e.parabolic-rectangular, are similar to that adoptedin BS 8110. The ultimate strain (εcu2) of the con-crete in compression is taken as 0.0035 in bothcodes but the position of the parabolic axis is fixedin EC 2 i.e. εc2 = 0.002 for fck ≤ 50 N mm−2,whereas in BS 8110 it is a function of thecompressive strength of concrete (Fig. 3.7).

The design compressive strength of concrete, fcd,is defined as

fcd = αcc fck/γC (8.10)

whereαcc coefficient taking account of long term

effects on the compressive strength = 0.85γC partial safety factor for concrete = 1.5

(Table 8.4)

Like BS 8110, EC 2 allows the use of an equiva-lent rectangular stress distribution for the designof cross-sections as shown in Fig. 8.4. This isslightly different to the rectangular distribution

Fig. 8.4 Simplified stress block for concrete at the ultimate limit state (based on Fig. 3.5, EC 2).

StressStrain

xλx

λ = 0.8 and η = 1 for fck ≤ 50 N mm–2

fcd = 0.85fck/γC

Note: If width of compression zonedecreases in direction of extremecompression fibre (e.g. circularcolumn), the value of ηfcd should bereduced by 10%

εcu2ηfcd

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324

Fig. 8.5 Design stress strain diagram for steel reinforcement (Fig 3.8, EC 2).

recommended in BS 8110, as the former is basedon cylinder rather than the cube strength of con-crete. Note that for fck ≤ 50 N mm−2 the depth ofcompression zone is 0.8x and the effective strengthis 0.85fck/1.5 (= 0.567fck) as η = 1.

8.6.2 REINFORCING STEELThe design steel stresses, fyd, are derived from theidealised (characteristic) stresses, fyk, by dividingby the partial safety factor for steel, γs:

f

fyd

yk

S

= γ

(8.11)

Figure 8.5 shows the idealised and design stress-strain curves for reinforcing steel recommended inEC 2. The idealised curve has an inclined topbranch. But for design either an inclined top branchcurve with a strain limit of εud (= 0.9εuk, see Table8.3) or a horizontal top branch curve with no limitto the steel strain and a maximum design stress offyk/γs can be used. The design equations which havebeen developed later in this chapter assume thestress-strain curve for steel has a horizontal topbranch.

8.7 Cover, fire, durability andbond (Cl 4, EC 2)

In EC 2 the cover to reinforcement is principally afunction of Fig. 8.6 Axis distance.

σ

εεuk

Es = 200 kN mm–2

fyd/Es

fyk

fyd = fyk/γs

Idealiseddesign

Idealisedcharacteristic

1. Fire resistance2. Durability3. Bond

Other factors include the quality of constructioncontrol and the quality of the surface against whichthe reinforcement is cast. EC 2 requirements in rela-tion to each of these aspects are discussed below.

8.7.1 FIREAs previously noted, fire design is covered inPart 1.2 of Eurocode 2 (EC 2-1-2). Exposure ofmembers to high temperatures in the event of afire can result in reduction of strength of both theconcrete and embedded steel reinforcement, as wellas spalling of the concrete cover. The fire resistanceof concrete members is principally related to thesize and shape of element as well as the cover tocentre of reinforcing bars i.e. axis distance (seeFig. 8.6). In the case of columns it is also a func-tion of the load supported. The risk of spalling is

a

asd

a = nominal axis distanceasd = axis distance to side of section

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325

Table 8.9 Minimum dimensions and axial distances to meet specified periods of fire resistance

Fire Beam One-way solid slab Braced column (6µfi = 0.7) −resistance 7Method A(min)

Simply supported 3Continuous Simply supported 3Continuous Exposed on Exposed on more1bmin/

2a (mm) 1bmin/a (mm) 4hs/a (mm) 4hs/a (mm) one side than one side5bmin/a (mm) 5bmin/a (mm)

R60 120/40 120/25 80/20 80/10 155/25 250/46160/35 200/12 350/40200/30300/25

R90 150/55 150/35 100/30 100/15 155/25 350/53200/45 250/25 8450/40300/40400/35

R120 200/65 200/45 120/40 200/20 175/35 8350/57240/60 300/35 8450/51300/55 450/35500/50 500/30

R240 280/90 280/75 175/65 280/40 295/70350/80 500/60500/75 650/60700/70 700/50

Notes:1bmin = beam width2a = axis distance3Applicable where specified detailing rules are observed and moment redistribution does not exceed 15%. Otherwise, consider assimply supported.4hs = depth of slab5bmin = column width6Ratio of the design axial load under fire conditions to the design resistance at normal temperature conditions conservativelytaken as 0.77Assumes (i) the effective length of the column under fire conditions l0,fi ≤ 3 m where l0,fi = 0.5 × actual length (l) forintermediate floors and 0.5l ≤ l0,fi ≤ 0.7l for the upper floor (ii) the first order end eccentricity under fire conditions, e = M0Ed,fi/N0Ed,fi ≤ 0.15b where M0Ed,fi and N0Ed,fi are the first order end moment and axial load under fire conditions and (iii) As ≤ 0.04Ac8Minimum of 8 bars required

Cover, fire, durability and bond

related to the type of aggregate used in the mix,with concrete made of siliceous aggregate morevulnerable to this form of damage.

EC 2-1-2 describes a number of methods forverifying the fire resistance of concrete membersincluding a method based on tabulated data whichis similar to the current approach recommendedin BS 8110. Table 8.9 shows values of minimumdimensions and associated axis distances for com-mon element types, necessary to achieve specifiedperiods of fire resistance. The tabulated values arevalid for normal weight concrete made of siliceousaggregate.

8.7.2 BOND AND DURABILITYThe nominal cover to reinforcement, cnom, isobtained from the minimum cover, cmin, by addingan allowance for likely deviations during construc-tion, ∆cdev i.e.

cnom = cmin + ∆cdev (8.12)

The recommended value of ∆cdev for rein-forced concrete is normally taken as 10 mm butfor concrete cast against uneven surfaces shouldbe increased by allowing larger deviations in design.According to clause 4.4.1.3, the minimum covershould be taken as 40 mm for concrete cast on

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326

Table 8.10 Minimum cover to reinforcement for durability, cmin,dur

Exposure class 1Minimum cover (mm)

X0 Not recommended for reinforced concrete

XC1 15 15 15 15 15 15 15 15XC2 – 25 25 25 25 25 25 25XC3/4 – 35 30 25 25 20 20 20

XD1 – – 235 230 30 225 25 25XD2 – – 340 335 235 330 30 30

XD3 – – – – 350 345 240 40XS1 – – – – 240 235 35 30XS2 – – 340 335 235 330 30 30XS3 – – – – – 350 245 45

Maximum free water/cement ratio 0.70 0.65 0.60 0.55 0.5 0.45 0.40 0.35Minimum cement content (kgm−3) 240 260 280 300 320 340 360 3804Lowest concrete class C20/25 C25/30 C28/35 C32/40 C35/45 C40/50 C45/55 C50/60

Notes:1Based on a design life of 50 years2Increase minimum cement content by 20 kgm−3

3Increase minimum cement content by 40 kgm−3 and reduce the free water/cement ratio by 0.054Concrete classes are for OPC concrete

sub-classes to which a structure could be subjectduring its design life. They are identical to thoseused in BS 8110 (see Table 3.5) and may occur singlyor in combination. As with BS 8110, the designermust turn to EN 206: Concrete – Performance,Production, Placing and Compliance Criteria and thecomplementary British Standard BS 8500: Part 1:Method of specifying and guidance for the specifier inorder to determine the minimum concrete qualityand minimum cover to reinforcement necessary toachieve the design life of the structure. Table 8.10which is based on advice in these documents givesfor particular exposure classes a summary of theminimum concrete quality (in terms of maximumwater/cement ratio and minimum cement content)and minimum covers to reinforcement necessaryto achieve a design life of 50 years for OPCconcrete. If blended mixes are used it is possibleto reduce the depth of concrete cover and/or thequality of the concrete indicated in Table 8.10 asdiscussed in section 3.8.1 of this book.

Finally it is worth noting that in clause 7.3.1 ofEC 2 it is also suggested that the presence of crackscan impair concrete durability and, like BS 8110,recommends that the maximum surface crack widthshould not generally exceed 0.3 mm in reinforcedconcrete members. This limiting crack width is

prepared ground (including blinding) and 65 mmwhere concrete is cast directly against soil.

The minimum cover to reinforcement is givenby the following expression

cmin = max{cmin,dur; cmin,b; 10 mm} (8.13)

wherecmin,dur is the minimum cover for durabilitycmin,b is the minimum cover for bond

From equation 8.13 it can be seen that theminimum cover should not be less than 10 mm.In clause 4.4.1.2(3) it is further recommendedthat in order to transmit bond forces safely andto ensure adequate compaction of the concrete,the minimum cover should not be less than cmin,b

where

cmin,b = diameter of bar, φ, provided the nominalmaximum aggregate size, dg ≤ 32 mm

The route to estimating cmin,dur is more complexand, like BS 8110, principally depends on

1. Environmental conditions2. Concrete quality

With regard to environmental conditions, EC 2defines six major exposure classes and eighteen

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327

8.3), provided moment redistribution does notexceed 20 per cent.

Fig. 8.7 shows the simplified stress blocks whichare used in BS 8110 and EC 2 to develop thedesign equations for bending. As previously noted,in EC 2 the maximum concrete compressive stressis taken as 0.85fck/1.5 and the depth of the com-pression block is taken as 0.8x for fck ≤ 50 N mm−2,where x is the depth of the neutral axis. The cor-responding values in BS 8110 are 0.67fcu/1.5 and0.9x as shown in Fig. 8.7(c).

The following sub-sections derive the equationsrelevant to the design of singly and doubly rein-forced beams according to EC 2. The correspondingequations in BS 8110 were derived in section3.9.1.1 to which the reader should refer for detailedexplanations and the notation used.

8.8.1.1 Singly reinforced beam design

(i) Ultimate moment of resistance (MRd)

MRd = Fcz (8.14)

Fc =

0 85. fck

1.50.8xb (8.15)

z = d − 0.4x (8.16)

Cl. 5.6.3 of EC 2 limits the depth of the neutralaxis (x) to 0.45d for concrete strength classes lessthan or equal to C50/60 (and 0.35d for concreteclasses C55/67 and greater) in order to provide aductile, i.e. under reinforced, section. Thus

x = 0.45d (8.17)

Note that the corresponding value for x in BS8110 is 0.5d.

Fig. 8.7 Singly reinforced section with rectangular stress block (a) Section (b) Strains (c) Stress Block (BS 8110)(d) Stress Block (EC 2)

Design of singly and doubly reinforced rectangular beams

achieved in practice by (a) providing a minimumarea of reinforcement and (b) limiting either themaximum bar spacing or the maximum bar diam-eter. These requirements are discussed individu-ally for beams, slabs and columns in sections 8.8.4,8.9.2 and 8.11.6, respectively.

8.8 Design of singly and doublyreinforced rectangular beams

8.8.1 BENDING (CL. 6.1, EC 2)When determining the ultimate moment of resistanceof concrete cross-sections, Cl. 6.1 of EC 2 recom-mends that the following assumptions are made:

1. plane sections remain plane2. the strain in bonded reinforcement is the same

as that in the surrounding concrete3. the tensile strength of concrete is ignored4. the compressive stresses in the concrete are

derived from the parabolic-rectangular stress–strain relationship shown in Fig. 8.3 or othersimplified stress–strain relationships, providedthey are effectively equivalent to Fig. 8.3, e.g. therectangular stress distribution shown in Fig. 8.4.

5. the stresses in the reinforcement are derived fromFig. 8.5.

6. the compressive strain in the concrete shouldnot exceed 0.0035.

These assumptions are similar to those listed inCl. 3.4.4.1 of BS 8110 with the possible omissionof the 0.95d limit on the lever arm i.e. z ≤ 0.95d.However, if this condition is observed, the steelstrain does not exceed 2.5 per cent which meansthat all classes of reinforcement are suitable (Table

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328

Combining equations (8.14), (8.15), (8.16) and(8.17) gives

MRd = 0.167fckbd2 (8.18)

c.f. Mu = 0.156fcubd 2 (BS 8110)

(ii) Area of tensile steel (As1)

M = Fsz (8.19)

F

fAs

yks

.=

1 15 1 (8.20)

A

Mf zs1yk

= .0 87

(8.21)

c.f. A

Mf zsy

= .0 87

(BS 8110)

Equation 8.21 can be used to calculate the areaof tension reinforcement provided that the designultimate moment, MEd ≤ MRd.

(iii) Lever arm (z)

M = Fcz

=

0 85. fck

1.5(0.8xb)z (from equation 8.15)

=

3 43.

fckb(d − z)z (from equation 8.16)

Solving for z gives

z = d[0.5 + ( . / . )0 25 3 3 4− Ko ] (8.22)

where K

Mf bdock

=2

c.f. z = d[0.5 + ( . / . )0 25 0 9− K ] (BS 8110)

where K

Mf bd

=cu

2

8.8.1.2 Doubly reinforced beamsIf the design ultimate moment is greater than theultimate moment of resistance i.e. MEd > MRd, thencompression reinforcement is required. Providedthat

d2/x ≤ 0.38 (i.e. compression steel has yielded)

whered2 is the depth of the compression steel from

the compression face and

x = (d − z)/0.4

the area of compression reinforcement, As2, isgiven by:

A

M Mf d ds2

Rd

yk

. ( )=

−−0 87 2

(8.23)

and the area of tension reinforcement, As1, is given by:

A

Mf z

As1Rd

yks2

. = +

0 87(8.24)

where z = d[0.5 + ( . / . )0 25 3 3 4− ′Ko ]Ko′ = 0.167

Equations 8.23 and 8.24 have been derived usingthe stress block shown in Fig. 8.8. This is similarto that used to derive the equations for the designof singly reinforced beams (Fig. 8.6d ) except forthe additional force due to the steel in the com-pression face.

d2

Fig. 8.8 Doubly reinforced section (a) Section (b) Strains (c) Stress Block (EC 2).

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329

Example 8.2 Bending reinforcement for a singly reinforced beam (EC 2)Determine the area of main steel, A s1, required for the beam assuming the following material strengths: fck = 25 Nmm−2 and fyk = 500 N mm−2

DESIGN MOMENT (M )

Total ultimate load (W) = (1.35gk + 1.5qk)span = (1.35 × 4 + 1.5 × 5)9 = 116.1 kN

Design moment (MEd) =

WB

8

116 1 9

8130 6

. . =

×= kNm


Ultimate load (w) = 1.35gk + 1.5qk

= 1.35 × 12 + 1.5 × 8 = 28.2 kNm−1

Design moment (MEd) =

wB 2 2

8

28 2 7

8172 7

. . =

×= kNm

Ultimate moment of resistance (MRd) = 0.167fckbd2

= 0.167 × 25 × 275 × 4502 × 10−6 = 232.5 kNm

Since MRd > MEd design as a singly reinforced beam

K

Mf bdo

Ed

ck

.

.= =

×× ×

=2

6

2

172 7 1025 275 450

0 124

z = d[0.5 + ( . / . )0 25 3 3 4− K o ]

= 450[0.5 + ( . . / . )0 25 3 0 124 3 4− × ] = 393.7 mm

A

M

f zs1

Ed

yk

mm .

.

. . = =

×× ×

=0 87

172 7 10

0 87 500 393 71008

62

Hence from Table 3.10, provide 4H20 (A s1 = 1260 mm2)

Example 8.3 Bending reinforcement for a doubly reinforced beam (EC 2)Design the bending reinforcement for the beam assuming the cover to the main steel is 40 mm, fck = 25 N mm−2 andfyk = 500 N mm−2

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330

Example 8.3 continuedULTIMATE MOMENT OF RESISTANCE (MRd)

Effective depthAssume diameter of main bar (φ) = 25 mmEffective depth (d) = h − c − φ/2 = 370 − 40 − 12.5 = 317 mm

Ultimate momentUltimate moment of resistance (MRd) = 0.167fckbd2 = 0.167 × 25 × 230 × 3172 × 10−6 = 96.5 kNmSince MRd < MEd design as a doubly reinforced beam.

COMPRESSION REINFORCEMENT (A s2)Assume diameter of compression bars (φ′) = 16 mmEffective depth (d2) = cover + φ′/2 = 40 + 16/2 = 48 mm

dx d2 48

0 4548

0 45 3170 34 0 38

.

. . .= =

×= < OK

Hence, A

M M

f d ds2

Ed Rd

yk

mm

. ( )

( . . )

. ( ) =

−−

=−

× −=

0 87

130 6 96 510

0 87 500 317 48291

2

62

Provide 2H16 (A s2 = 402 mm2 from Table 3.10)

TENSION REINFORCEMENT (A s1)z = d[0.5 + ( . / . )0 25 3 3 4− K o ] = d[0.5 + ( . . / . )0 25 3 0 167 3 4− × ] = 0.82d

Hence, A

M

f zAs1

Rd

yks2 mm

.

.

. . = + =

×× × ×

+ =0 87

96 5 10

0 87 500 0 82 317291 1144

62

Provide 3H25 (A s1 = 1470 mm2 from Table 3.10).

EC 2 it can vary between 21.8° and 45° and itis this feature which is responsible for possiblereductions in the volume of shear reinforcement asillustrated in Fig. 8.10. Here it can be seen that fora given design shear strength, V, EC 2 generallyrequires fewer links than BS 8110.

It is worth noting, however, that using a reducedvalue of strut angle also results in a concomitantincrease in tensile force in the bottom chord (BT)which necessitates that longitudinal reinforcementextends further in the span than required forbending alone. Thus, any potential savings in thevolume of shear reinforcement has to be offset

2H16

3H25

d2 = 48

8.8.2 SHEAR (CL. 6.2, EC 2)EC 2 uses the so called variable strut inclinationmethod for shear design. It is slightly more com-plex than the procedure in BS 8110 but can resultin savings in the amount of shear reinforcementrequired. Like BS 8110, EC 2 models shearbehaviour on the truss analogy in which the concreteacts as the diagonal struts (shown in broken lineand labelled DC in Fig. 8.9), the stirrups act as thevertical ties, VT, the tension reinforcement formsthe bottom chord, BT and the compression steel/concrete forms the top chord, TC. Whereas in BS8110 the strut angle, θ, has a fixed value of 45° in

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331

BS 8110

BS 8110: θ = 45° V = Vc + Vs

Minimum links

cot θ = 1

cot θ = 2.5

VRd max

Asfyd/s

EC 2

Vc

V > V* EC 2 requires fewerlinks than BS 8110

V Shear strength,V

Test results

Fig. 8.9 (a) Beam and reinforcement (b) analogous truss.


VT

DC

Concretestrut

Tensionreinforcement

Stirrups

z

BT

TC

VTDC

θ

Fig. 8.10 Influence of links on shear strength.

against increases in the volume of longitudinalreinforcement. Other major differences in designprocedure are:

1. EC 2 compares shear forces rather than shearstresses.

2. The maximum shear capacity of concrete isnot a fixed value (i.e. the lesser of 0.8

fcu or

5 N mm−2 as in BS 8110) but depends on theapplied shear.

3. Where the applied shear exceeds the shearresistance of the concrete, the shear reinforce-ment should be capable of resisting all the shearacting on the section.

EC 2 identifies four basic shear forces for designpurposes, namely VEd, VRd,c, VRd,max and VRd,s. VEd

is the design shear force and is a function of theapplied loading. It is normally determined at a dis-tance, d, from the face of supports (Fig. 8.11).

(b)

(a)

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332

VRd,c is the design shear resistance of the memberwithout shear reinforcement and is given by:

VRd,c = [CRd,ck(100ρ1 fck)1/3 + k1σcp]bwd

≥ (νmin + k1σcp)bwd (8.25)

whereCRd,c = 0.18/γc

k = 1 +

200d

< 2.0 with the effective depth,

d in mm

ρ1 =

Ab d

sl

w

< 0.02

in whichAsl is the area of the tensile reinforcement,

which extends ≥ (lbd + d) beyond thesection considered

bw is the smallest width of the cross-section inthe tensile area

νmin = 0.035k3/2f ck1/2

k1 = 0.15σcp = NEd/Ac < 0.2fcd

in whichNEd is the axial force in the cross-sectionAc is the cross-sectional area of concrete

Expression 8.25 is empirical and not too dissim-ilar to the expression which appears at the bottomof Table 3.8 of BS 8110 and is used to determinethe design concrete shear stress (Table 3.8).

If VEd < VRd,c no shear reinforcement is requiredexcept, possibly, in beams where it is normal toprovide a minimum amount of shear links. How-ever, if VEd > VRd,c shear failure may occur as aresult of either compressive failure of the diagonalconcrete strut or diagonal tension failure of themember (Fig. 3.21).

The concrete strut capacity, VRd,max, is given by:

VRd,max = bwzν1 fcd /(cot θ + tan θ) (8.26)

wherez ≈ 0.9dfcd = αcc fck/γm = 0.85fck/1.5 (for fck ≤ 50 N/mm2)

(Note: αcc = 1.0 may be used here.)ν1 = 0.6(1 − fck/250) for fck ≤ 50 N/mm2

θ is the angle between the concrete strut andthe axis of the beam (Fig. 8.9)

According to EC 2, the strut angle (and hencethe strut capacity) is not unique but can varybetween cot θ = 2.5 (i.e. 21.8°) and cot θ = 1 (i.e.45°), depending on the value of the applied shear.Using the identity

10 5 2

cot tan sin cos . sin

θ θθ θ θ

+= =

expression 8.26 can be transposed to make theconcrete strut angle the subject of the formula asfollows

θ . sin

( / ). ( / )

=−

−0 50 153 1 250

1 V b df fRd,max w

ck ck

(8.27)

Expression 8.27 can be used to calculate theminimum strut angle for a given value of appliedshear by equating VRd,max = VEd subject to the con-dition that cot θ lies between 1.0 and 2.5.

As noted above, if VEd > VRd,c shear reinforcementmust be provided. Provided VEd ≤ VRd,max, the areaof shear reinforcement can be estimated from thefollowing expression by equating VEd = VRd,s

V

As

zfRd,ssw

ywd cot = θ (8.28)

whereVRd,s is the shear resistance of the member

governed by ‘failure’ of the stirrupsAsw is the cross-sectional area of the shear

reinforcements is the spacing of shear reinforcement

(see 8.8.2.2)fywd is the design yield strength of the shear

reinforcement.

Expressions 8.26 and 8.28 are not empirical butcan be derived via the analogous truss, as follows.Consider a reinforced concrete beam with stirrupsuniformly spaced at a distance s, subject to adesign shear force VEd as shown in Fig. 8.12. Theresulting strut angle is θ. Assuming the height ofthe analogous truss is z, the number of links perstrut is equal to z cot θ/s and, hence, the shearresistance of the member, controlled by yielding ofthe shear reinforcement (i.e. leading to ‘failure’ ofVT), VRd,s, is given by

Fig. 8.11

45°

d

udl load

Forces in this region are assumed tobe transmitted to the support bycompression

d

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333

VEd

z

Stirrups

ss

z cot θ

θBT

TC

VTDC z

(shear span)

Fig. 8.12

Fig. 8.13

a c

bV

VV/sin θ

θ

Fig. 8.14


a c

b

BT

DC

bWz cos θ

θ

z

VRd,s = number of links in shear span × totalcross-sectional area × design stress

= [(z cot θ)/s] × Asw × fywd

=

As

zfswywd cot θ (equation 6.8, EC 2)

From Fig. 8.13 it can be seen that the concretestruts are bw wide and (z cos θ) deep. Hence, thecompressive capacity of each strut, DC, is given by

DC = cross-sectional area of strut × designstrength of concrete cracked in shear

= bw(z cos θ) × (ν1 fcd)

Shear failure of the strut will not occur providedDC equals or exceeds the design shear force actingon the strut equal to V/sin θ (Fig. 8.14), i.e.

Vsin θ

≤ DC = (ν1 fcd)bw(z cos θ)

Hence the shear resistance of the member con-trolled by crushing of compression struts, VRd,max,is given by

VRd,max = DC sin θ = (ν1 fcd)bw(z cos θ) sin θ

Using the identity cos2θ + sin2θ = 1 gives

VRd,max = (ν1 fcd)bwz cos θ sin θ /(cos2θ + sin2θ)

Dividing top and bottom by cos θ sin θ andsimplifying gives

=

+ ( )

cos sin cos sin (cos sin )/(cos sin )

νθ θ θ θ

θ θ θ θ1 2 2f b zcd w

/

=

+ ( )

cot tan ν

θ θ11

f b zcd w (equation 6.9, EC 2)

From Fig. 8.14 it can be seen that for equilib-rium a tensile force of V/tan θ also acts on thesection. Assuming that half of this force acts inthe bottom chord, the additional tensile forcepresent in the longitudinal reinforcement, ∆Ftd, isgiven by

∆Ftd = 0.5VEd /tan θ (8.29)

The presence of this additional longitudinal forceis responsible for the shift rule discussed in sec-tion 8.8.4.4, necessitating that longitudinal tensionsteel extend further in the span than required forbending alone. Also since the tensile force in thelongitudinal steel due to bending is MEd/z, the totalforce in the tension reinforcement, Fs, is given by

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334

Fs = Force due to bending+ Force due to shear = MEd/z + ∆Ftd

Its value should not exceed MEd,max whereMEd,max is the maximum moment within the hog-ging or sagging region that contains the sectionconsidered.

The design process can be summarised as follows:

1. Calculate the design shear force, VEd.2. Determine the design shear resistance of the

member without shear reinforcement, VRd,c.3. If VEd < VRd,c shear reinforcement can be

omitted except in beams where a minimumarea of shear reinforcement must be provided(see 8.8.2.1).

4. If VEd > VRd,c all the shear force must be resistedby the shear reinforcement. Provided VEd <VRd,max, the area of shear reinforcement canbe determined using expression 8.28. VRd,max isestimated from (8.26) assuming initially cot θ= 2.5. However, if the result is VEd > VRd,max

a larger strut angle, θ, may be used. Themaximum allowable value of θ is 45° (i.e. cot θ= 1). The minimum value of the strut anglefor a given design shear force, VEd, can bedetermined from (8.27) and used, in turn, tocalculate the area of shear reinforcement from(8.28).

5. If the strut angle exceeds 45°, however, a deeperconcrete section or higher concrete strength mustbe provided and steps (2)–(4) repeated.

8.8.2.1 Shear reinforcement areas (ρw)(Cl. 9.2.2, EC 2)

As noted above, minimum reinforcement should beprovided in beams where VEd < VRd,c. The minimumshear reinforcement ratio, ρw, is obtained from

ρw = Asw /(sbw sin α) ≥ ρw,min (8.30)

whereAsw is the area of shear reinforcement within

length ss is the spacing of the shear reinforcementbw is the breadth of the memberα = 90° for vertical stirrups ⇒ sin α = 1ρw,min is the minimum area of shear reinforcement

= (0.08 fck )/fyk (8.31)

8.8.2.2 Spacing of shear reinforcement(Cl. 9.2.2, EC 2)

EC 2 recommends that the maximum longitudinalspacing of shear reinforcement (sl,max) should notexceed 0.75d(1 + cot α)

whereα = 90° for vertical stirrups ⇒ cot α = 0

Example 8.4 Design of shear reinforcement for a beam (EC 2)Design the shear reinforcement for the beam in Example 8.2.

DESIGN SHEAR FORCE, VEd

This should be determined at distance d from the face of the support but for simplicity has been calculated at thecentre of supports.

VEd = 0.5(ωB) = 0.5 × 28.2 × 7 = 98.7 kN

SHEAR RESISTANCE OF CONCRETE (VRd,c)

fck = 25 N mm−2

CRd,c = 0.18/γC = 0.18/1.5 = 0.12 N mm−2

k

d . .= + = + = <1

2001

200450

1 67 2 0 OK

Assuming all the tension reinforcement is taken onto supports and anchored

ρ1

1260275 450

0 0102 0 02

. .= =×

= <A

b dsl

w

OK

σcp = 0

νmin = 0.035k3/2f ck1/2 = 0.035 × 1.673/2 × 251/2 = 0.378

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335



VRd,c = [CRd,ck(100ρ1fck)1/3 + k1σcp]bwd

= [0.12 × 1.67(100 × 0.0102 × 25)1/3]275 × 450

= 72 994 N ≥ (νmin + k1σcp)bwd = 0.378 × 275 × 450 = 46 778 N

Since VRd,c < VEd, shear reinforcement must be provided.

COMPRESSION CAPACITY OF COMPRESSION STRUT, VRD,MAX, ASSUMING θ = 21.8°ν1 = 0.6(1 − fck/250) = 0.6(1 − 25/250) = 0.54

fcd = αccfck/γc = 0.85 × 25/1.5 = 14.2 N mm−2 (Note αcc = 1.0 may be used.)

VRd,max = bwzν1fcd/(cot θ + tan θ)

= [275 × (0.9 × 450) × 0.54 × 14.2/(2.5 + 0.4)] × 10−3 = 294.5 kN > VEd OK

DIAMETER AND SPACING OF LINKS(i) Where VEd < VRd,c, provide minimum shear reinforcement, ρw,min, according to

ρw,min = (0.08 fck )/fyk = (0.08 25 )/500 = 8 × 10−4

ρw,min = A sw/sbw sin α⇒ Asw/s = 8 × 10−4 × 275 × 1 = 0.22 mm (assuming the use of vertical links)

Maximum spacing of links, smax, is

smax = 0.75d = 0.75 × 450 = 338 mm

Hence, from Table 3.13 provide H8 at 300 mm centres, A sw/s = 0.335 mm.

(ii) Where VEd > VRd,c provide shear reinforcement according to:

V

As

zfRd,ssw

ywd = cot θ = 98700 N

⇒

Assw

( . )( / . ) . .=

× ×=

987000 9 450 500 1 15 2 5

0 224

Hence provide H8 at 300 mm centres (A sw /s = 0.335) throughout.

Example 8.5 Design of shear reinforcement at beam support (EC 2)Design the shear reinforcement for the beam shown in Fig. 8.15, assuming it resists an ultimate shear force atdistance d from the face of the support of 450 kN. The characteristic material strengths are fck = 25 N mm−2 andfywv = 500 N mm−2.

Fig. 8.15

450 kN

4H25

b = 300 mm

d = 500 mmd

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336

SHEAR RESISTANCE OF CONCRETE (VRd,c)

fck = 25 N mm−2

CRd,c = 0.18/γC = 0.18/1.5 = 0.12 N mm−2

k

d . .= + = + = <1

2001

200

5001 63 2 0 OK

ρ1

1960300 500

0 013 0 02

. .= =×

= <A

b dsl

w

OK

σcp = 0

νmin = 0.035k3/2f ck1/2 = 0.035 × 1.633/2 × 251/2 = 0.364


= [0.12 × 1.63(100 × 0.013 × 25)1/3]300 × 500 × 10−3

= 93.6 kN ≥ (νmin + k1σcp)bwd = 0.364 × 300 × 500 × 10−3 = 54.6 kN

Since VRd,c < VEd = 450 kN, shear reinforcement must be provided.

CHECK COMPRESSION CAPACITY OF COMPRESSION STRUT, VRd,max, ASSUMING θ = 21.8°ν1 = 0.6(1 − fck /250) = 0.6(1 − 25/250) = 0.54

fcd = αccfck/γc = 0.85 × 25/1.5 = 14.2 N mm−2 (Note αcc = 1.0 may be used.)

VRd,max = bwzν1fcd/(cot θ + tan θ)

= [300 × (0.9 × 500) × 0.54 × 14.2/(2.5 + 0.4)] × 10−3 = 357 kN

Since VRd,max < VEd, strut angle > 21.8°. From equation 8.27

θ . sin

( / ). ( / )

. sin( / ). ( / )

.=−

=× ×× −

= °− −0 50 153 1 250

0 5450 10 300 500

0 153 251 25 25030 31 1

3V b df fRd,max w

ck ck

DIAMETER AND SPACING OF LINKSProvide shear reinforcement according to:

V

A

szfRd,s

swywd kN cot = =θ 450

⇒

Assw

( . )( / . ) cot . .=

× × °=

4500000 9 500 500 1 15 30 3

1 344


smax = 0.75d = 0.75 × 500 = 375 mm

Therefore H12 at 150 mm centres (A sw/s = 1.507) would be suitable.Note. If longitudinal reinforcement is fully anchored at support and VEd = 450 kN ≤ 0.5bwdνfcd = 0.5 × 300 × 500

× 0.54 × 14.2 × 10−3 = 575 kN (OK), shear force may be reduced to (av/2d) × 450 = 225 kN. In this case θ = 21.8°is OK and shear reinforcement is reduced by over 50 per cent.


9780415467193_C08 9/3/09, 2:55 PM336

337

Table 8.11 Basic span/depth ratios for reinforced concrete members

Structural system K Highly stressed concrete Lightly stressed concreteρ = 1.5% ρ = 0.5%

Simply supported beams and slabs 1.0 14 20End span of continuous beams and slabs 1.3 18 26Interior spans of continuous beams and slabs 1.5 20 30Cantilever beams and slabs 0.4 6 8

B

dK f f . .

/

= +

+ −

11 1 5 3 2 13 2

cko

ckoρ

ρρρ

if ρ ≤ ρo (8.32)

B

dK f f .

/

= +− ′

+ ′

11 1 51

12

1 2

cko

cko

ρρ ρ

ρρ

if ρ > ρo (8.33)

wherel/d is the limiting span/depth ratioK is the factor to take into account the

different structural systems, given in Table 8.11ρo = fck × −10 3

ρ is the required tension reinforcement ratioρ′ is the required compression reinforcement

ratio

The values in Table 8.11 assume the steel stressat the critical section, σs, is 310 N mm−2, corres-ponding roughly to the stress under characteristicload when fyk = 500 N mm−2. Where other steelstresses are used, the values in the table can bemultiplied by 310/σs. It will normally be conserv-ative to assume that

σ

νs

yk s,req

s,pro

=310

500

f AA

(8.34)

whereAs,req is the area of steel requiredAs,prov is the area of steels provided

For flanged sections where the ratio of the flangebreadth to the rib width, beff/bw, exceeds 3, the val-ues in the table should be multiplied by 0.8. Lin-ear interpolation may be used to estimate values ofl/d for intermediate values of beff/bw. For beamsand slabs, other than flat slabs (see Fig. 3.52), where


8.8.3 DEFLECTIONCrack and deflection control are the two mainserviceability checks required by EC 2. EC 2 alsoincludes a check on the stress in the concrete andsteel reinforcement under specified loadings but pastperformance of concrete structures would seem tosuggest that this check is almost certainly unneces-sary in the UK and is therefore not discussed.

Like BS 8110, EC 2 recommends that surfacecrack widths should not generally exceed 0.3 mmin order to avoid poor durability. As previouslymentioned, this is normally achieved by observingthe rules on minimum steel areas and either themaximum bar spacing or the maximum bar dia-meter. The rules for beams are presented in section8.8.4.

EC 2 requirements for deflection control appearto be less onerous than those in BS 8110, namely

1. the calculated sag of a beam, slab or cantileverunder quasi-permanent loading should notexceed span/250.

2. the deflection occurring after construction of theelement should not exceed span/500.

It is true that the creep component of deflectionis related to quasi-permanent loads and if appear-ance is the only criterion it would be reasonable tobase the deflection check on this load combination.However, if function, including possible damage tofinishes, cladding and partitions is of concern, thenthe use of characteristic load would seem moreappropriate.

Although it is possible to estimate actual deflec-tions of concrete elements in order to check com-pliance it is more practical to use limiting span/depth ratios as in BS 8110. Table 8.11 shows basicspan/depth ratio for commonly occurring reinforcedconcrete members and support conditions. Theyhave been obtained using equations 8.32 and8.33.

9780415467193_C08 9/3/09, 2:55 PM337


338

(a) Non-continuous member (b) Continuous member

ln ln

a i = lesser of h/2 and t /2

h

t

a i = lesser of h/2and t /2

t

h

span lengths exceed 7 m, the values of l/d shouldbe multiplied by 7/leff.

whereleff effective span of the member = ln + a1 + a2

(Cl. 5.3.2.2)

in whichln is the clear distance between the faces of

supportsa1,a2 are the distances at each end of the span

shown in Fig. 8.16

Fig. 8.16 Definition of a1 and a2 (a) Non-continuous member (b) Continuous member (based on Fig. 5.4, EC 2).

Example 8.6 Deflection check for concrete beams (EC 2)Carry out a deflection check for Examples 8.2 and 8.3.

FOR EXAMPLE 8.2fck = 25 N mm−2 ⇒ ρo = fck × 10−3 = 25 × 10−3 = 5 × 10−3

ρ

. = =

×= × −A

bdsl 1008

275 4508 15 10 3

Since ρ > ρo, use equation 8.33

B

dK f f .

. .

.

./

= +− ′

+ ′

= +×× −

=

−

−11 1 5

112

1 0 11 1 5 255 10

8 15 10 015 6

123

3cko

cko

ρρ ρ

ρρ

σ

νs

yk s,req

s,pro

Nmm

= =

× ××

= −310

500

310 500 1008

500 1260248 2

f A

AAllowable span/depth ratio = basic l/d × 310/σs = 15.6 × 310/248 = 19.5Actual span/depth ratio = 7000/450 = 15.6 < allowable OK

FOR EXAMPLE 8.3

fck = 25 N mm−2 ⇒ ρo = fck × 10−3 = 25 × 10−3 = 5 × 10−3

ρ

. = =

×= × −A

bds1 1144

230 31715 69 10 3

′ = =

×= × −ρ

.

Abd

s2 291230 317

3 99 10 3

9780415467193_C08 9/3/09, 2:56 PM338

339

Since ρ > ρo,

B

dK f f .

/

= +− ′

+ ′

11 1 51

12

1 2

cko

cko

ρρ ρ

ρρ

= +×

× − ×

+×

×

=−

− −

−

− . .

. .

.

./

1 0 11 1 5 255 10

15 69 10 3 99 101

1225

3 99 105 10

14 53

3 3

3

3

1 2

σ

νs

yk s,req

s,pro

Nmm

= =

× ××

= −310

500

310 500 1144

500 1470241 2

f A

AAllowable span/depth ratio = basic l/d × 310/σs × 7/span = 14.5 × 310/241 × 7/9 = 18.6Actual span/depth ratio = 9000/317 = 28.4 > allowable Not OK


8.8.4 REINFORCEMENT DETAILS FOR BEAMSThis section outlines EC 2 requirements regardingthe detailing of beams with respect to:

1. reinforcement percentages2. spacing of reinforcing bars3. anchorage lengths4. curtailment of reinforcement5. lap lengths.

For a brief discussion on each of these aspectsrefer to section 3.9.1.6 of this book.

8.8.4.1 Reinforcement percentages(Cl. 9.2.1.1, EC 2)The cross-sectional area of the longitudinal tensilereinforcement, As1, in beams should not be lessthan the following:

A

ff

b d b dsctm

ykt t1 0 26 0 0013 . .= ≥

wherebt is the breadth of sectiond effective depthfctm mean tensile strength of concrete

(see Table 8.2)fyk characteristic yield stress of reinforcement

(= 500 N mm−2)

The area of the tension, A s1, and of the com-pression reinforcement, A s2, should not be greaterthan the following other than at laps:

A s1, A s2 ≤ 0.04Ac

whereAc is the cross-sectional area of concrete.

8.8.4.2 Spacing of reinforcement (Cl. 8.2, EC 2)The clear horizontal or vertical distance betweenreinforcing bars should not be less than the following:

(a) maximum bar diameter(b) maximum aggregate size, dg + 5 mm(c) 20 mm

To ensure that the maximum crack width doesnot exceed 0.3 mm EC 2 recommends that theminimum reinforcement necessary to control crack-ing should be provided and either the maximumbar spacing or the maximum bar diameter shouldnot exceed the values shown in Table 8.12. Thesteel stress, σs, can conservatively be estimatedusing equation 8.35.

σ

γ νs

yk s,req

s s,pro

=f mA

nA(8.35)

wherefyk = 500 N mm−2

γs = 1.15m quasi-permanent load (= gk + Ψ2qk)n design ultimate load (= γGgk + γQqk)As,req area of steel requiredAs,prov area of steel provided

8.8.4.3 Anchorage (Cl. 8.4.3, EC 2)EC 2 distinguishes between basic and design anch-orage lengths. The basic anchorage length, lb,rqd, isthe length of bar required to resist the maximumforce in the reinforcement, As.σsd. Assuming a con-stant bond stress of fbd the basic anchorage lengthof a straight bar is given by:

lb,rqd = (φ/4)(σsd/fbd) (8.36)


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340

Hatched zone: ‘poor’ bond conditions

300

250

Direction of concreting

Direction of concretingDirection of concreting

h < 250

h > 600h > 250

(a) (b) (c)

Table 8.12 Maximum bar spacing and bar diameter for crack control (based on Table 7.2N and7.3N, EC 2)

Steel Stress (N mm−2) Maximum bar spacing (mm) Maximum bar size (mm) φ s

wk = 0.3 mm wk = 0.3 mm

160 300 32200 250 25240 200 16280 150 12320 100 10360 50 8

Note: The values in the table assume cnom = 25 mm and fck = 30 N mm−2

Table 8.13 Ultimate bond stress, fbd (N mm−2), assuming good bond conditions

fck (N mm−2) 12 16 20 25 30 35 40 45 50

fbd where φ ≤ 32 mm 1.65 2.00 2.32 2.69 3.04 3.37 3.68 3.98 4.35

Fig. 8.17 Definition of bond conditions (Fig. 8.2, EC 2).

whereφ diameter of bar to be anchoredσsd design stress of the bar at the position from

where the anchorage is measuredfbd ultimate bond stress = 2.25η1η2 fctd (8.37)

whereη1 coefficient related to the quality of the bond

between the concrete and steel. EC 2 statesthat bond conditions are considered to begood for:(a) all bars in members with an overall

depth (h) of less than or equal to250 mm (Fig. 8.17a)

(b) all bars located within the lower250 mm depth in members with anoverall depth exceeding 250 mm(Fig. 8.17b)

(c) all bars located at a depth greater thanor equal to 300 mm in members withan overall depth exceeding 600 mm(Fig. 8.17c).

η1 = 1.0 when good bond exists and 0.7 for allother conditions

η2 is related to bar diameter and is 1.0 forφ ≤ 32 mm and (132 − φ)/100 for φ > 32 mm

fctd is the design value of the concrete tensilestrength = αct fctk,0.05/γC

in whichαct = 1.0fctk,0.05 is the 5% fractile characteristic axial

tensile strength of concrete (see Table 8.2)

Table 8.13 shows values of the ultimate bondstress for a range of concrete strengths assuming

9780415467193_C08 11/3/09, 11:13 AM340

341

good bond conditions and φ ≤ 32 mm. When bondconditions are poor the values in the table shouldbe multiplied by 0.7.

The design anchorage length, lbd, is used to deter-mine the curtailment of longitudinal reinforcementin beams as discussed in section 8.8.4.4 and isgiven by:

lbd = α1α2α3α4α5lb,rqd ≥ lb,min (8.38)

whereα1–α5 are coefficients which take account of,

respectively, shape of bar, concretecover/spacing between bars, transversereinforcement not welded to the mainreinforcement, transverse reinforcementwelded to the main reinforcement andpressure transverse to the reinforcement.The product of α1–α5 can conservativelybe taken as 1.

lb,min minimum anchorage length given by:lb,min = maximum of {0.3.lb,rqd; 10φ;100 mm} (for anchorages in tension)lb,min = maximum of {0.6.lb,rqd; 10φ;100 mm} (for anchorages in compression)

As a simplification for the shapes shown inFig. 8.18(b) to (d) an equivalent anchorage length,lb,eq, may be used where lb,eq is given by

lb,eq = α1lb,rqd (8.39)

in whichα1 = 1.0 for straight and curved bars in

compression= 1.0 for straight bars in tension= 0.7 for curved bars in tension if cd > 3.0φ

wherecd is as shown in Fig. 8.19 for straight and

curved barsφ is the diameter of the bar to be anchored

Fig. 8.18 Common methods of anchorage. Equivalentanchorage lengths for bends, hooks and loops (based onFig. 8.1, EC 2)

Table 8.14 Design anchorage lengths, lbd, as multiples of bar diameters assuming φ ≤ 32 mm andgood bond conditions for a range of concrete strengths

Steel Grade Concrete Class

C20/25 C25/30 C30/37 C35/45 C40/50 C45/55 C50/60

Deformed bars Straight bars in tension 47 41 36 33 30 28 25type 2 or compressionfyk = 500 N mm−2 Curved bars in tension* 33 29 25 23 21 20 18

Curved bars in 47 41 36 33 30 28 25compression

*In the anchorage region, cd > 3φ


Bbdφ

Bb

Bb.eq

Bb.eqBb.eq

(a) Basic tension anchoragelength, Bb, for any shapemeasured along thecentre line

(b) Bend (d) Loop

(c) Hook

Table 8.14 shows the design anchorage lengthsfor straight and curved bars as multiples of bardiameters for high yield bars ( fyk = 500 N mm−2)embedded in a range of concrete strengths. Notethat the values in the table apply to good bond

(a) cd = lesser of a/2, c1 and c (b) cd = lesser of a/2 and c1

c1

a a

c

c1

Fig. 8.19 Values of cd for beams and slabs (a) straightbars (b) bent or hooked bars (based on Fig. 8.3, EC 2).

9780415467193_C08 9/3/09, 2:57 PM341


342

conditions and to bar sizes less than or equal to 32mm. Where bond conditions are poor the values inthe table should be divided by 0.7 and where thebar diameter exceeds 32 mm the values shouldbe divided by [(132 − φ)/100]. The minimum radiito which reinforcement may be bent is shown inTable 8.15 whilst the anchorage lengths for straights,hooks, bends and loops are shown in figure 8.18.

8.8.4.4 Curtailment of bars (Cl. 9.2.1.3, EC 2)The curtailment length of bars in beams is ob-tained from Fig. 8.20. The theoretical cut-off pointis based on the design bending moment curve whichhas been horizontally displaced by an amount al

in the direction of decreasing bending moment, aconsequence of the additional longitudinal force inthe tension steel due to the applied shear force.This is sometimes referred to as the ‘shift rule’.It is also necessary to ensure that the bar extendsan anchorage length beyond the point where itis fully required for bending and shear. Thus inrelation to the bending moment envelope, barshould extend (a1 + lbd) beyond the point where

fully required and a1 beyond the point where nolonger required.

If the shear resistance is calculated according tothe ‘variable strut inclination method’ the shift, a l,is given by:

a1 =

z2

(cot θ − cot α) (8.40)

whereθ is the angle of the concrete strut to the

longitudinal axisα is the angle of the shear reinforcement to the

longitudinal axis.z ≈ 0.9d

Example 8.7 Calculation of anchorage lengths (EC 2)Calculate the anchorage lengths for straight and curved bars in tension as multiples of bar size assuming:

1. The bars are high yield of diameter, φ ≤ 32 mm2. Concrete strength class is C20/253. Bond conditions are good4. cd > 3φ.

STRAIGHT BARSHigh yield bars, fyk = 500 N mm−2

Concrete strength, fck = 20 N mm−2

Ultimate bond stress, fbd = 2.32 (Table 8.13)Design strength of bars, fyd = fyk/γS = 500/1.15

The basic anchorage length, lb,rqd, is

lb,rqd = (φ/4)(fyd/fbd)

= (φ/4)(500/1.15)/2.32 ≈ 47φHence, the anchorage length, lbd, is

lbd = α1α2α3α4α5lb,rqd = 1.0 × 47φ = 47φ (see Table 8.14)

CURVED BARSThe calculation is essentially the same for this case except that α1 = 0.7 for curved bars and therefore, lbd, is

lbd = lb,eq = α1lb,rqd = 0.7 × 47φ ≈ 33φ (see Table 8.14)

Table 8.15 Minimum diameter of bends, hooksand loops (Table 8.1N, EC 2)

Bar diameter Minimum mandrel diameter

φ ≤ 16 mm 4φφ > 16 mm 7φ

9780415467193_C08 9/3/09, 2:58 PM342

343

For members with vertical links al will be equalto:

a1 =

z2

(cot θ − cot 90°) =

z2

cot θ

Since, in practice, the strut angle often takes thevalue cot θ = 2.5 the shift, a1, is given by

a1 = 1.25z

EC 2 also recommends the following rules forthe curtailment of reinforcement in beams subjectedto predominantly uniformly distributed loads.

(i) Top reinforcement at end supports. Inmonolithic construction, even where simple sup-ports have been assumed in design, the section atsupports should be designed for a bending momentarising from partial fixity of at least 25 per cent ofthe maximum bending moment in the span.

(ii) Bottom reinforcement at end supports.When there is little or no fixity at end supports,at least 25 per cent of the reinforcement providedat mid-span should be taken into the support andanchored as shown in Fig. 8.21. The tensile forceto be anchored, FE, is given by:

FE = |VEd|.a1/z + NEd (8.41)

whereVEd is the shear force at the end supporta1 is obtained from equation 8.40NEd is the axial force in the member

(iii) Bottom reinforcement at intermediatesupports. At least 25 per cent of the reinforcementrequired at mid-span should extend to intermedi-ate supports. The anchorage length should not beless than 10φ for straight bars.

The reinforcement required to resist possiblepositive moments due to, for example, accidentalactions such as explosions and collisions, shouldbe continuous which may be achieved by means oflapped bars as shown in Fig 8.22a or 8.22b.

Fig. 8.23 shows the simplified rules for curtail-ment of reinforcement in simply supported andcontinuous beams recommended in The ConcreteCentre publication ‘How to design concrete struc-tures using Eurocode 2’.

8.8.4.5 Lap lengths (Cl. 8.7.3, EC 2)Lap length, lo, is given by:

lo = α1α2α3α5α6lb,rqd ≥ lo,min = 0.3α6lb,rqd (8.42)

- Envelope of MEdlz + NEdA

A

B

C

Ibd

- acting tensile force FsB - resisting tensile force FRsC

Ibd

IbdIbd

Ibd

a1

a1

Ibd

Ibd

Ibd

Ftd

Ftd

Fig. 8.20 Envelope line for curtailment of reinforcement in flexural members (based on Fig. 9.2, EC 2).


Fig. 8.21 Curtailments and anchorages of bottomreinforcement at end supports (based on Fig. 9.3, EC 2).

(a) (b)

100%25%

b

lbd ≥ 10φ

lbd ≥ 10φ

0.08l

25% 100%

0.08l

9780415467193_C08 9/3/09, 2:58 PM343


344

Fig. 8.22 Anchorage at intermediate supports (Fig. 9.4, EC 2).

l ≥ 10φ

φ φ

lbd lbd

wherelb,rqd anchorage lengthlo,min minimum lap length which should be not

less than 15φ or 200 mmα1–α5 are as for equation 8.38α6 = (ρ1/25)0.5: 1.0 ≤ α6 ≤ 1.5 (Table 8.15)

in whichρ1 percentage of reinforcement lapped within

0.65lo from the centre of the lap lengthconsidered

Table 8.16 shows the compression and tensionlap lengths as multiples of bar size for grade 500bars ( fyk = 500 N mm−2) embedded in a rangeof concrete strengths. Note that the values in thetable apply to good bond conditions and to barsizes less than or equal to 32 mm. Where the bondconditions are poor the values in the table shouldbe divided by 0.7 and where the bar diameterexceeds 32 mm the values should be divided by[(132 − φ)/100].

(a)

(b)

30% 100%60% 30%100%

0.3l – a l

0.15l + a l ≥ lbd

0.3l + a l

Face of support

25% 100%

lbd

0.08l

Fig. 8.23 Simplified detailing rules for beams (a) simpleend supports (b) internal supports of continuous beams.

Table 8.16 Lap lengths as multiples of bar size for grade 500 ( fyk = 500 N mm−2) deformed bars

% of lapped bars α6 Concrete classrelative to the totalx-section area, ρ1 C20/25 C25/30 C30/37 C35/45 C40/45

Compression < 25% 1 47 41 36 33 30& tension laps 33% 1.15 54 48 42 38 35

50% 1.4 66 58 51 47 42> 50% 1.5 71 62 54 50 45

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345

gk = 25.5 kN m–1

qk = 20 kNm–1

b = 300 mm

h = 600 mm

6 m

Example 8.8 Design of a simply supported beam (EC 2)Design the longitudinal and link reinforcement for the beam shown in Fig. 8.24 assuming Class XC1 exposure. Thebeam is made of class C25/30 concrete of bulk density 25 kNm−3 and grade 500 steel.

Fig. 8.24

DESIGN MOMENT (M )LoadingPermanentSelf weight of beam = 0.6 × 0.3 × 25 = 4.5 kNm−1

Total permanent load, gk = 4.5 + 25.5 = 30 kNm−1

VariableTotal variable load, qk = 20 kNm−1

Ultimate loadTotal ultimate load, ω = 1.35gk + 1.5qk = 1.35 × 30 + 1.5 × 20 = 70.5 kNm−1

Design moment

MEd kNm

. . = =

×=

ωB 2 2

8

70 5 6

8317 25

ULTIMATE MOMENT OF RESISTANCE, MRd

Effective depthMinimum cover, cmin = max{cmin,dur; cmin,b; 10 mm}

For Class XC1 exposure, cmin,dur = 15 mm (Table 8.10)Assume diameter of longitudinal reinforcement is 25 mm, hence cmin,b = 25 mmNominal cover to main steel = cmin,b + ∆cdev = 25 + 10 = 35 mm

Assuming diameter of links = 8 mm, nominal cover to links = 35 − 8 = 27 mm > nominal cover for durability (= 15+ 10 = 25 mm). Hence, effective depth, d, is

d = 600 − 27 − 8 − 25/2 = 552 mm

[Note that, in practice, spacers in multiples of 5 mm are fixed to links. Nominal cover to links will be either 25 mmor 30 mm (to ensure at least 35 mm to main bars).]

Ultimate momentMRd = 0.167fcubd2 = 0.167 × 25 × 300 × 5522 × 10−6 = 381.6 kNmSince MRd > MEd design as a singly reinforced beam.


9780415467193_C08 9/3/09, 2:59 PM345


346

Example 8.8 continuedMAIN STEEL, AS1

K

Mf bdo

ck

= =×

× ×=

.

.2

6

2

317 25 1025 300 552

0 139

z = d[0.5 + ( . / . )0 25 3 3 4− K o ]

= 552[0.5 + ( . . / . )0 25 3 0 139 3 4− × ] = 473 mm

A

M

f zs

yk

mm1

62

0 87

317 25 10

0 87 500 4731542

.

.

. = =

×× ×

=

Therefore, from Table 3.10 provide 4H25 (A s1 = 1960 mm2)

CHECK MAXIMUM AND MINIMUM STEEL AREAS AND MINIMUM BAR SPACING

0.04Ac ≥ A s ≥ 0 26.

ff

b dctm

ykt ≥ 0.0013btd

0.04Ac = 0.04 × bh = 0.04 × 300 × 600 = 7200 mm2 > A s1 OK

0 26 0 26

0 3 25

500300 552 221

2 32. .

.

/f

fb dctm

ykt mm=

×= × =

≥ 0.0013b td = 0.0013 × 300 × 552 = 215 mm2 < A s1 OK

Clear distance between bars = 1/3[b − (2 × cover) − 4φ]

= 1/3[300 − 2 × 35 − 4 × 25] = 43 mm

> min{φ; dg (say 20 mm) + 5 mm; 20 mm] OK

SHEAR REINFORCEMENT

W

VEd

VEd

RBRA

Design shear force, VEd

VEd = 0.5(ωB) = 0.5 × 70.5 × 6 = 211.5 kN

Shear resistance of concrete (VRd,c)fck = 25 N mm−2

CRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2

k

d . .= + = + = <1

2001

200552

1 6 2 0 OK

9780415467193_C08 9/3/09, 2:59 PM346

347


ρ1

1960300 552

0 0118 0 02

. .= =×

= <A

b dsl

w

OK

σcp = 0

νmin = 0.035k3/2f ck1/2 = 0.035 × 1.63/2 × 251/2 = 0.354 N mm−2

VRd,c = [CRd,ck (100ρ1fck)1/3 + k1σcp]bwd

= [0.12 × 1.6(100 × 0.0118 × 25)1/3]300 × 552

= 98243 N ≥ (νmin + k1σcp)bwd = 0.354 × 300 × 552 = 58 622 N OK

Since VRd,c < VEd, shear reinforcement must be provided.

Check compression capacity of compression strut, VRd,max, assuming ~ = 21.8°ν = 0.6(1 − fck/250) = 0.6(1 − 25/250) = 0.54

fcd = αccfck/γc = 0.85 × 25/1.5 = 14.2 N mm−2

VRd,max = bwzνfcd/(cot θ + tan θ)

= [300 × (0.9 × 552) × 0.54 × 14.2 / (2.5 + 0.4)] × 10−3

= 394.2 kN > VEd OK

Diameter and spacing of links


Where VEd < VRd,c, provide minimum shear reinforcement, ρw,min, according to

ρw,min = (0.08 fck )/fyk = (0.08 25)/500 = 8 × 10−4

ρw,min = A sw/sbwsin α

⇒ A sw/s = 8 × 10−4 × 300 × 1 = 0.24


smax = 0.75d = 0.75 × 552 = 414 mm

Hence, from Table 3.13 provide H8 at 300 mm centres, A sw /s = 0.335

VRd,s = 0.335 × 497 × (500/1.15) × 2.5 × 10−3 = 181 kN

Where VEd > 181 kN provide shear reinforcement according to:

V

A

szfRd,s

swywd N cot = =θ 211500

VE

d

3 m 3 m

VRd,s (with minimum links) = 181 kN

x

x

VE

d =

2115

00 N

x = VRd,s/ω = 181/70.5 = 2.567 m

9780415467193_C08 9/3/09, 2:59 PM347


348

Example 8.8 continuedHowever, it can be seen that VEd = 181 kN occurs at x = 2.567 m which is less than d from the face of the support,and there is therefore no need to reduce the link spacing beyond this point.

DEFLECTION

fck = 25 N mm−2 ⇒ ρo = fck × = × = ×− − −10 25 10 5 103 3 3

ρ

. = =

×= × −A

bds1 1542

300 5529 3 10 3

Since ρ > ρo,

B

dK f f .

. .

.

/

= +− ′

+ ′

= +×

× −

=

−

−11 1 5

112

1 0 11 1 5 255 10

9 3 10 015

123

3cko

cko

ρρ ρ

ρρ

σ

νs

yk s,req

s,pro

Nmm

. = =

× ××

= −310

500

310 500 1542

500 1960243 9 2

f A

AAllowable span/depth ratio = basic 1/d × 310/σs = 15 × 310/243.9 = 19Actual span/depth ratio = 6000/552 = 10.9 < allowable OK

REINFORCEMENT DETAILSThe sketches below show the main reinforcement requirements for the beam. For reasons of buildability the actualreinforcement details may well be slightly different.

300 mm300 mm

20H8-300

FE = VEdal/z = (VEdcot θ)/2 = 1.25VEd for cot θ = 2.5⇒FE = 1.25 × 211.5 = 264.4 kN

As,req = (264.4 × 103)/(500/1.15) = 608 mm2

lbd = (41 × 25) × 608/1960 = 318 mm measured from front face of support to end of bar ≥ 0.3lbd,req = 0.3 × (41 × 25) = 308 mm OK

2H12

H8 links

4H25

300 mm

552 mm

48 mm

2H12

H8 links

H25

125 mm

152.5

87.55700 mm

9780415467193_C08 9/3/09, 3:00 PM348

349

x = 0.616d

200 kNmm–2

εS = 0.002175 = εy

fS = 500/1.15

d

εy = 0.002175

εcu = 0.0035

Example 8.9 Analysis of a singly reinforced beam (EC 2)Calculate the maximum variable load that the beam shown below can carry assuming that the load is (i) uniformlydistributed or (ii) occurs as a point load at mid-span.

MAXIMUM UNIFORMLY DISTRIBUTED LOAD, qK

Permanent load, gk = 0.5 × 0.3 × 25 = 3.75 kNm−1

Total ultimate load, ω = 1.35gk + 1.5qk

= 1.35 × 3.75 + 1.5 × qk


MOMENT CAPACITY OF SECTIONEffective depth, d, is

d = h − cover − φ/2 = 500 − 40 − 25/2 = 447 mm

K

Mf bd

Mo

ck

= =× ×

2 225 300 447

z = d[0.5 + ( . / . )0 25 3 3 4− K o ]

= 447[0.5 + ( . / . )0 25 3 3 4 25 300 4472− × × × ×M ] (1)

A

M

f z

M

zs

yk

mm .

.

= =× ×

=0 87 0 87 500

1960 2 (2)

Solving equations (1) and (2) simultaneously gives

M = 295.6 kNm and z = 346.7 mm

⇒ x

d zd

.

.

. . .=

−=

−= ≤

0 4

447 346 7

0 4250 7 0 616mm (to ensure section is under-reinforced)

= 0.616 × 447 = 275 mm OK

fck = 25 Nmm–2

b = 300

h = 500

40 mm cover

4H25 (As1 = 1960 mm2)7 m

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350

Example 8.9 continuedSubstituting into

M =

ωB2

8

295 6

1 35 3 75 1 5 78

2

. ( . . . )

=× + ×qk

Hence

qk = 28.8 kNm−1

MAXIMUM POINT LOAD, QKFactored permanent load, WP = (1.35gk) span

= (1.35 × 3.75) 7 = 35.44 kN

Factored variable load, Wv = 1.5Q k

Substituting into

M

W W = +P VB B

8 4

295 6

35 44 78

1 5 74

. .

.

=×

+×Qk

Hence Q k = 100.8 kN

8.9 Design of one-way solid slabs

8.9.1 DEPTH, BENDING, SHEAREC 2 requires a slightly different approach to BS8110 for designing one-way spanning solid slabs.To calculate the depth of the slab, the designer mustfirst estimate the percentage of steel required inthe slab for bending. Generally, most slabs will belightly reinforced, i.e. ρ < 0.5%. The percentage ofreinforcement can be used, together with the sup-port conditions, to calculate an appropriate span/depth ratio using equations 8.32 and 8.33. Theeffective depth of the slab is obtained by multiplyingthis ratio by the span (see Example 8.10). The actualarea of steel required in the slab for bending iscalculated using the equations developed in section8.8.1. Provided this agrees with the assumed value,the calculated depth of the slab is acceptable.

The designer will also need to check that theslab will not fail in shear. The shear resistance ofthe slab can be calculated as for beams (see section8.8.2). According to Cl. 6.2.1(3), where the designshear force (VEd) is less than the design shearresistance of the concrete alone (VRd,c) no shearreinforcement is necessary. The same requirementappears in BS 8110.

Where VEd > VRd,c, all the shear force must beresisted by the shear reinforcement. The actual area

of shear reinforcement in slabs can be calculatedas for beams (section 8.8.2). This should not beless than the minimum value, ρw,min, given by

ρw,min = (0.08 fck )/fyk

For members with vertical shear reinforcement, themaximum longitudinal spacing of successive seriesof shear links, smax, should not exceed

smax = 0.75d

Moreover, the maximum transverse spacing of linkreinforcement should not exceed 1.5d. Cl. 9.3.2(1)also notes that shear reinforcement should not beprovided in slabs less than 200 mm deep.

8.9.2 REINFORCEMENT DETAILS FOR SOLIDSLABS

This section outlines EC 2 requirements regardingthe detailing of slabs with respect to:

1. reinforcement percentages2. spacing of reinforcement3. anchorage and curtailment of reinforcement4. crack control

1. Reinforcement areas (Cl. 5.4.2.1.1, EC 2)In EC 2, the maximum and minimum percentagesof longitudinal steel permitted in beams and slabsare the same, namely:

9780415467193_C08 9/3/09, 3:01 PM350

351

(a)

(b)

Face of support

15% 100%

lbd

0.15l, ≥ Ibd + d

the adjacent span as shown in Fig. 8.25 and beanchored. At an end support the moment to beresisted may be reduced to 15 per cent of the max-imum moment in the adjacent span.

The Concrete Centre publication ‘How to designconcrete structures using Eurocode 2’ recommendsthat the simplified curtailment rules for continuousslabs shown in Fig. 8.26 should be used.

A

ff

b d b dsctm

ykt t1 0 26 0 0013 . .≥ ≥

As1, As2 ≤ 0.04Ac

whereb breadth of sectiond effective depthAc cross-sectional area of concrete (bh)fyk characteristic yield stress of reinforcement

The area of transverse or secondary reinforce-ment should not generally be less than 20 per centof the principal reinforcement, i.e.:

As (trans) ≥ 0.2 × As (main)

2. Spacing of reinforcementThe clear distance between reinforcing bar shouldnot be less than the following:

(a) maximum bar diameter(b) 20 mm(c) dg + 5 mm where dg is the maximum aggregate

size.

The maximum bar spacing in slabs, smax, for themain reinforcement should not exceed the following:

smax = 3h ≤ 400 mm generally and 2h ≤ 250 mmin areas of maximum moment

where h denotes the overall depth of the slab.The maximum bar spacing in slabs, smax, for the

secondary reinforcement should not exceed thefollowing:

smax = 3.5h ≤ 450 mm generally and3h ≤ 400 mm in areas of maximum moment

3. Anchorage and curtailment (Cl 9.2.1.3 & 9.3.1.2)For detailing the main reinforcement in slabs,similar provisions to those outlined earlier for beamsapply.

Cl. 9.3.1.2 recommends that where end supportsare simply supported, 50 per cent of the calculatedspan reinforcement should continue up to the sup-port and be anchored to resist a force, FE, given by

FE = |VEd|.a1/z + NEd

where |VEd|, z and NEd are as defined for equation8.41 and a1 is equal to d if the slab is not rein-forced for shear or is given by equation 8.40 ifshear reinforcement is provided.

If partial fixity exists along an edge of a slabtop reinforcement should be provided capable ofresisting at least 25 per cent of the maximummoment in the adjacent span. This reinforcementshould extend at least 0.2 times the length of

Design of one-way solid slabs

Fig. 8.26 Simplified curtailment rules for slabs (a) simplesupport (b) continuous member.

0.2l

100%25%

Fig. 8.25

Section A-A

A

A

25%

15% 25%

100%

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352

4. Crack widths (Cl. 7.3.3, EC 2)According to EC 2, where the overall depth ofthe slab does not exceed 200 mm and the codeprovisions with regard to reinforcement areas, spac-ing of reinforcement, anchorage and curtailmentof bars, etc., discussed above have been applied,no further measures specifically to control crackingare necessary.

Furthermore, where at least the minimum rein-forcement area has been provided, the limitationof crack width to less than 0.3 mm for reinforcedconcrete may generally be achieved by either limit-ing the maximum bar spacing or maximum bardiameter in accordance with the values given inTable 8.12.

DETERMINE EFFECTIVE DEPTH OF SLAB AND AREA OF MAIN STEELEstimate overall depth of slabAssume reinforcement ratio, ρ, is 0.35%. From equation 8.32, basic span/effective depth ratio for a simply supportedslab is approximately 31.

Minimum effective depth, d = = ≈

span

basic ratiomm

4650

31150

Hence, take d = 155 mm

Minimum cover, cmin = max{cmin,dur; cmin,b; 10 mm}

For Class XC1 exposure, cmin,dur = 15 mm (Table 8.10)Assume diameter of main steel, Φ = 10 mm, therefore cmin,b = 10 mmNominal cover to reinforcement, cnom = cmin + ∆cdev = 15 + 10 = 25 mmOverall depth of slab (h) = d + Φ/2 + cnom = 155 + 10/2 + 25 = 185 mm

LoadingPermanent

Self weight of slab (gk) = 0.185 × 25(kNm−3) = 4.625 kNm−2

VariableTotal variable load (qk) = 4 kNm−2

Ultimate loadFor 1m width of slab total ultimate load = (1.35g k + 1.5qk)span

= (1.35 × 4.625 + 1.5 × 4)4.65 = 56.93 kN

Example 8.10 Design of a one-way spanning floor (EC 2)Design the floor shown in Fig. 8.27 for an imposed load of 4 kN mm−2, assuming the following material strengths:fck = 30 Nmm−2, fyk = 500 Nmm−2. The environmental conditions fall within exposure class XC1.

Fig. 8.27

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353



Design momentDesign moment,

M

WlEd kNm

. . . = =

×8

56 93 4 65

833 1

Ultimate momentUltimate moment of resistance (MRd) = 0.167fckbd 2

= 0.167 × 30 × 1000 × 1552 × 10−6 = 120 kNm

Since MRd > MEd no compression reinforcement is required.

Main reinforcement (As1)

K

M

f bdo

ck

.

.= =

×× ×

=2

6

2

33 1 10

30 1000 1550 046

z = d(0.5 + ( . / . )0 25 3 3 4− K o ) ≤ 0.95d = 0.95 × 155 = 147.3 mm

= 155(0.5 + ( . . / . )0 25 3 0 046 3 4− × ) = 155 × 0.957

A

M

f zs

yk

mm1

62

0 87

33 1 10

0 87 500 147 3517

.

.

. . = =

×× ×

=

Hence from Table 3.22, H10 at 150 mm centres (A s1 = 523 mm2 m−1) would be suitable.The required reinforcement ratio, ρ, in this case is

ρ

. . %= =

×= =

A

bds1 517

1000 1550 00334 0 334

Check estimated effective depth of slab

Design service stress, σ

νs yk

req

pro

Nmm . ≈ × × = × × = −5

8

5

8500

517

523308 9 2f

A

A

Modification factor =

310 310

308 91 0

σ s

.

.= =

Basic span/effective depth ratio is given by

B

dK f f . .

/

= +

+ −

11 1 5 3 2 1

3 2

cko

ckoρ

ρρρ

= +××

+ ×

××

−

=−

−

−

− . .

. .

. .

/

1 0 11 1 5 3030 10

3 34 103 2 30

30 10

3 34 101 33 4

3

3

3

3

3 2

Modified span/effective depth ratio = basic ratio × modification factor

= 26.4 × 1.0 = 33.4

Actual span/effective depth ratio = =

4650

15530 < allowable OK

Therefore, use a slab with an effective depth = 155 mm, overall depth = 185 mm and main steel = H10 at 150 mmcentres.

Check bar spacing and steel areaMaximum bar spacing for main steel < 3h = 3 × 185 = 555 mm ≤ 250 mm (in areas of maximum moment) OKMaximum area of steel, A s,max = 0.04Ac = 0.04 × 185 × 103 = 7400 mm2 m−1 OK

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354

Example 8.10 continuedMinimum reinforcement area,

A

ff

b d b dsctm

ykt t,min . .= ≥0 26 0 0013

=

×× = − .

.

/

0 260 3 30

50010 155 233

2 33 2 1mm m

≥ 0.0013bd = 0.0013 × 103 × 155 = 202 mm2 m−1 > A s1 OK

SECONDARY REINFORCEMENTProvide H8 at 300 mm centres (A s (tras) = 168 mm2 m−1)

A s (trans) ≥ 0.2A s (main) = 0.2 × 523 = 105 mm2 m−1 OK

Maximum bar spacing < 3.5h = 3.5 × 185 = 648 mm ≤ 400 mm (in areas of maximum moment) OK

SHEAR REINFORCEMENT

VEd

W

RA RB

VEd

4.5 m

Ultimate load (W ) = 51.7 kN

Design shear force (VEd)

V

WEd kN= = =

. .

2

56 93

228 5

Shear resistance of concrete alone, VRd,c

fck = 30 N mm−2

CRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2

k

d . .= + = + = ≤1

2001

200155

2 1 2 0

Assuming 50 per cent of mid-span steel does not extend to the supports A s1 = 260 mm2 m−1. Hence

ρ1 3

26010 155

0 00168 0 02

. .= =×

= <A

b ds1

w

OK

σcp = 0

νmin = 0.035k3/2f ck1/2 = 0.035 × 23/2 × 301/2 = 0.542

VRd,c = [CRd,ck (100ρ1fck)1/3 + k1σcp]bwd

= [0.12 × 2(100 × 0.00168 × 30)1/3]103 × 155

= 63780 N ≥ (νmin + k1σcp)bwd = 0.542 × 103 × 155 = 84010 N

Since VRd,c (= 84 kN) > VEd (28.5 kN) no shear reinforcement is required which is normally the case for slabs.

9780415467193_C08 9/3/09, 3:02 PM354

355

16-H08-300

FE = VEd(d/z) = 28.5/0.9 = 31.7 kN

As,req = (31.7 × 103)/(500/1.15) = 73 mm2

lbd = (36 × 10) × 73/260 = 101 mm measured from front face of support to end of bar ≥ 0.3lb,min = 0.3 × (36 × 10) = 108 mm

30mm

0801-H10-150

155mm

08

01

Curtail 50% of bars x mm in from centreline of both supports wherex = 0.15l – al(= d ) = 0.15 × 4650 – 155 = 542 mm

01 01

16-H08-300

01-H10-150

01-H10-150 542

108

150

Example 8.10 continuedREINFORCEMENT DETAILSThe sketches below show the main reinforcement requirements for the slab.

Reinforcement areas and bar spacingMinimum reinforcement areas and bar spacing rules were checked above and found to be satisfactory.

Crack widthSince the overall depth of the slab does not exceed 200 mm and the rest of the code provision have been met, nofurther measures specifically to control cracking are necessary.

fck = 25 N mm–2

fyk = 500 Nmm–2

ρc = 25 kNm–3


Example 8.11 Analysis of one-way spanning floor (EC 2)Calculate the maximum uniformly distributed variable load that the floor shown below can carry.

9780415467193_C08 9/3/09, 3:02 PM355


356

Example 8.11 continuedEFFECTIVE SPANFrom Cl 5.3.2.2. of EC 2, effective span of slab, leff, is

leff = ln + a1 + a2

whereln clear distance between the face of supportsa1, a2 are the lesser of half the width of the support, t/2, (= 150/2 = 75 mm) or half the overall depth of the slab,

h/2 (= 150/2 = 75 mm) at the ends of the slab

⇒ leff mm= + + = 2850

150

2

150

23000

VARIABLE LOAD CAPACITYThis can only really be estimated by trial and error. Try qk = 11.4 kNm−1.

Check bendingSelf weight of slab (gk) = 0.15 × 25 kNm−3 = 3.75 kNm−2

Ultimate load (ω) = 1.35gk + 1.5qk

= 1.35 × 3.75 + 1.5 × 11.4

Design moment, M

g q

( . . )

( . . . . ) . = =

× +=

× + ×=

ωB B2 2 2

8

1 35 1 5

8

1 35 3 75 1 5 11 4 3

824 9k k kNm

Effective depth of slab (d) = h − cover − Φ/2 = 150 − 25 − 10/2 = 120 mm

K

Mf bdo

ck

.

.= =

×× ×

=2

6

2

24 9 1025 1000 120

0 0692

z d K ( . ( . / . )) ( . ( . . / . )) = + − = + − × =0 5 0 25 3 3 4 120 0 5 0 25 3 0 0692 3 4 112o mm

A

M

f zs

y

mm provided mm1

62 2

0 87

24 9 10

0 87 500 112511 628

.

.

. = =

×× ×

= < =

Check deflection

Design service stress, σ s yk

s,req

s,prov

Nmm . ≈ × × = × × = −5

8

5

8500

511

628254 3 2f

A

A

Modification factor =

310 310254 3

1 22σs

.

.= =

ρ ρ

. . = =

×= = × < = × = ×− − −A

bdfs1

o ck511

1000 1200 00426 4 26 10 10 25 103 3 3

The basic span/effective depth ratio is given by

B

dK f f . .

/

= +

+ −

11 1 5 3 2 1

3 2

cko

ckoρ

ρρρ

= +××

+

××

−

=−

−

−

− . .

. .

.

/

1 0 11 1 5 2525 10

4 26 103 2 25

25 10

4 26 101 21

3

3

3

3

3 2

Allowable span/effective depth ratio = basic ratio × modification factor

= 21 × 1.22 = 25.6 > actual (3000/120 = 25) OK

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357

Example 8.11 continuedCheck shear capacityDesign shear force, VEd1, is

VEd kNm

( . . . . ) . = =

× + ×=

ωB

2

1 35 3 75 1 5 11 4 3

233 24

Shear resistance of concrete alone, VRd,c

fck = 25 N mm−2

CRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2

k

d . .= + = + = ≤1

2001

200120

2 3 2 0

Assuming all bars are taken onto supports

ρ1 3

62810 120

0 00523 0 02

. .= =×

= <A

b ds1

w

OK

σcp = 0

νmin = 0.035k3/2f ck1/2 = 0.035 × 23/2 × 251/2 = 0.495


= [0.12 × 2(100 × 0.00523 × 25)1/3]103 × 120

= 67848 N ≥ (νmin + k1σcp)bwd = 0.495 × 103 × 120 = 59400 N

VRd,c (= 67.85 kN) > VEd (= 33.24 kN) OK

Thus the slab is capable of supporting a uniformly distributed variable load of 11.4 kNm−2.

8.10 Design of pad foundationsThe design of pad foundations was discussed insection 3.11.2.1 and found to be similar to that forslabs in respect of bending. However, the designermust also check that the pad will not fail due toface, transverse or punching shear. EC 2 require-ments in respect of these three modes of failure arediscussed below.

8.10.1 FACE SHEAR (CL. 6.4.5, EC 2)According to Cl. 6.4.5, adjacent to the column theapplied ultimate shear stress, νEd, should not exceedthe maximum allowable shear resistance, νRd,max

where νEd and νRd,max are given by, respectively,equations 8.43 and 8.44:

ν βEd

Ed

0

=Vu d

(see Fig. 3.75, Chapter 3) (8.43)

whereVEd applied shear forced mean effective depth of the slab, which may

be taken as (dy + d z)/2 where dy, d z effective

depth of section in the y and z directionsrespectively

u0 length of column peripheryβ = 1 where no eccentricity of loading exists

νRd,max = 0.5νfcd (8.44)

where

ν = 0.6[1 − ( fck/250)]

f

fcd cc

ck

c

= αγ

8.10.2 TRANSVERSE SHEAR (CL. 6.4.4, EC 2)The critical section for transverse shear failure nor-mally occurs a distance d from the face of the col-umn. Shear reinforcement is not necessary providedthe ultimate design shear stress, νEd, is less that thedesign shear resistance, νRd where νEd and νRd aregiven by, respectively, equations 8.45 and 8.46:

νEd

Ed,red =V

ud(8.45)

Design of pad foundations

9780415467193_C08 11/3/09, 11:15 AM357


358

whereu is the perimeter being consideredd is the mean effective depth of the sectionVEd,red = VEd − ∆VEd

in whichVEd is the applied shear force∆VEd is the net upward force within the control

perimeter

The design shear resistance, νRd,c, is given by

νRd = CRd,ck(100ρ1 fck)1/3 × (2d/a)

≥ νmin × (2d/a) (8.46)

whereCRd,c = 0.18/γC

k = 1 +

200d

≤ 2.0

fck characteristic concrete strengthνmin = 0.035k3/2f ck

1/2

d is the effective depth of the sectiona is the distance from the periphery of the

column to the control perimeter considered

ρ ρ ρ1 1 1 0 02 .= =

×

≤y zsl,y

y

sl,z

z

Abd

Abd

in whichρ1y, ρ1z tension steel ratio in the y and z

directionsdy, dz effective depth of section in the y and z

directions

8.10.3 PUNCHING SHEAR (CL. 6.4.3 & 6.4.4, EC 2)No punching shear reinforcement is required ifthe applied shear stress, νEd, is less than the designpunching shear resistance of concrete, νRd, whereνEd and νRd are calculated using, respectively, equa-tions 8.47 and 8.48:

ν βEd

Ed =Vu d1

(8.47)

whereVEd ultimate design shear force. For a

foundation this is normally calculated alongthe perimeter of the base of the truncatedpunching shear cone, assumed to form at26.6° (Fig. 8.28)

u1 is the basic control perimeter (Fig. 8.29)β is a coefficient which takes account of the

effects of eccentricity of loading. In caseswhere no eccentricity of loading is possible,β may be taken as 1.0.

2d

2d

2d 2d

Fig. 8.29 Critical perimeters (Fig. 6.13, EC 2).

Fig. 8.28 Design model for punching shear at ULS(Fig. 6.16, EC 2).

Loaded area

θ

θ ≥ tan–1 (1/2)

d

Pad foundation

νRd = [CRd,ck(100ρfck)1/3 + k1σcp] × (2d/a)

≥ [νmin + k1σcp] × (2d/a) (8.48)

where

CRd,c = 0.18/γC

k = 1 +

200d

≤ 2.0

k1 = 0.1fck characteristic concrete strengthνmin = 0.035k3/2f ck

1/2

ρ ρ ρ .= =

×

≤1 1 0 02y zsl,y

y

sl,z

z

Abd

Abd

in whichρ1y, ρ1z tension steel ratio in the y and z

directionsdy, dz effective depth of section in the y and z

directions

9780415467193_C08 9/3/09, 3:04 PM358

359

Where νEd exceeds νRd, shear reinforcement willneed to be provided such that νEd ≤ νRd,cs

whereνRd,cs is the design value of punching shear

resistance of slab with shear reinforcementcalculated in accordance with Cl. 6.4.5 ofEC 2.

σcp = (σcy + σcz)/2

in whichσcy, σcz respectively, NEdy/Acy and NEdz/Acz

whereNEd longitudinal forceAc area of concrete

Design of pad foundations

Shear failure could arise:

(a) at the face of the column(b) at a distance d from the face of the column(c) punching failure of the slab

FACE SHEARLoad on footing due to column is 1.35 × 900 + 1.5 × 300 = 1665 kNDesign shear stress at the column perimeter, νEd, is

ν βEd

Ed

o

Nmm .

( ) . = = ×

×× ×

= −V

u d1 0

1665 10

4 350 5302 24

32

where d = h − cover − diameter of bar = 600 − 50 − 20 = 530 mm. Maximum punching shear resistance at thecolumn perimeter, νRd,max, is

νRd,max = 0.5νfcd = 0.5 × 0.528 × 20 = 5.28 N mm−2 > νEd OK

where

H20 at 250c/ceach way

Example 8.12 Design of a pad foundation (EC 2)The pad footing shown below supports a column which is subject to axial characteristic permanent and variableactions of 900 kN and 300 kN respectively. Assuming the following material strengths, check the suitability of thedesign in shear.

fck = 30 N mm−2

fyk = 500 N mm−2

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360

Example 8.12 continuedν = 0.6[1 − (fck/250)] = 0.6[1 − (30/250) ] = 0.528

f

fcd cc

ck

c

Nmm . .

= = × = −αγ

1 030

1 520 2

TRANSVERSE SHEAR

From above, design shear force on footing, VEd = 1665 kN

Earth pressure, p

VE

Ed

base areakNm = = = −1665

3185

22

Ultimate load on shaded area is

∆VEd = pE(3 × [3 − 0.795]) = 1223.8 kN

Applied shear force is

VEd,red = VEd − ∆VEd = 1665 − 1223.8 = 441.2 kN

Design transverse shear stress, νEd, is

νEd

Ed,red Nmm .

. = =

××

= −V

bd

441 2 10

3000 5300 28

32

where b = width of footing = 3 m

fck = 30 N mm−2

CRd,c = 0.18/γc = 0.18/1.5 = 0.12 N mm−2

k

d . .= + = + = <1

2001

200530

1 61 2 0 OK

ρ ρ ρ1 1 1 3 3

126010 530

126010 530

0 00238 0 02

. .= = × =×

××

= <y xsl,y sl,xA

bdAbd

OK

σcp = 0

νmin = 0.035k3/2f ck1/2 = 0.035 × 1.613/2 × 301/2 = 0.39

Design shear resistance of concrete, νRd,c, is given by

νRd,c = [CRd,ck(100ρ1fck)1/3 + k1σcp] × (2d/a)

= [0.12 × 1.61(100 × 0.00238 × 30)1/3 + 0] × 2

= 0.74 N mm−2 ≥ [νmin + k1σcp] × (2d/a) = 0.39 × 2 = 0.78 Nmm−2

Since νEd (= 0.28 Nmm−2) < νRd,c (= 0.78 Nmm−2) no shear reinforcement is required.

530mm

795 mm

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361

Example 8.12 continuedPUNCHING SHEAR

Design of columns

lateral forces in that plane. Thus braced columnsare assumed to not contribute to the overall hor-izontal stability of a structure and as such are onlydesigned to resist axial load and bending due tovertical loading. For further information on bracedcolumns refer to section 3.13.3. The design ofbraced columns involves consideration of the follow-ing aspects which are discussed individually below:

(a) slenderness ratio, λ(b) threshold slenderness, λlim

(c) first order effects(d) second order moments(e) reinforcement details.

8.11.1 SLENDERNESS RATIO (CL. 5.8.3.2, EC 2)In EC 2 the slenderness ratio, λ, is defined as:

λ =

lio (8.49)

wherelo is the effective length of the columni is the radius of gyration of the uncracked

concrete section

Note that in BS 8110 the slenderness ratio isbased on the section depth (b or h) and not itsradius of gyration.

8.11 Design of columnsColumn design is largely covered within sections5.8 and 6.1 of EC 2. The design procedures in EC2 are perhaps slightly more complex than thoseused in BS 8110, although the final result is simi-lar in both codes.

Column design generally involves determiningthe slenderness ratio, λ, of the member. If it liesbelow a critical value, λlim, the column can simplybe designed to resist the axial action and momentobtained from an elastic analysis, but including theeffect of geometric imperfections. These are termedfirst order effects. However, when the column slen-derness exceeds the critical value, additional (secondorder) moments caused by structural deformationscan occur and must also be taken into account. InEC 2, the slenderness ratio above which columnsare subjected to second order effects has to beevaluated while in BS 8110 it is taken as 15 forbraced columns and 10 for unbraced columns.

In this chapter only the design of the most com-mon types of columns found in building structures,namely braced columns, will be described. A col-umn may be considered to be braced in a givenplane if the bracing element or system (e.g. coreor shear walls) is sufficiently stiff to resist all the

Check punching shear at 2d from face of column. Basic control perimeter, u1, is

u1 = column perimeter + 2π(2d)

= 4 × 350 + 2π(2 × 530) = 8060 mm

Area within critical perimeter, A = 4 × 350(2 × 530) + 3502 + π(2 × 530)2 = 5.14 × 106 mm2

Ultimate load on shaded area, ∆VEd = pE × A = 185 × 5.14 = 950.9 kN

Applied shear force, VEd,red = VEd − ∆VEd = 1665 − 950.9 = 714.1 kN

Punching shear stress, νEd

Ed,red .

= =××

Vu d1

3714 1 108060 530

= 0.17 Nmm−2 < νRd,c = 0.39 Nmm−2 OK

Hence, no shear reinforcement is required.

2d

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362

background paper to the UK National Annex toEC 2, PD 6687, recommends that when calculat-ing the effective height of a column in which thestiffness of adjacent columns do not vary signific-antly, k1 and k2 should be calculated ignoring thecontribution of the attached columns. Moreover,the contribution of the attached beams should bemodelled as 2(EI/L), irrespective of end conditionsat their remote ends, to allow for the effect ofcracking. Calibration exercises suggest that use ofthese recommendations will lead to effective lengthswhich are similar to those currently obtained usingthe coefficients in Table 3.19 of BS 8110 (repro-duced as Table 3.27), and which could thereforebe used as an alternative method of estimating theeffective lengths of braced columns, if desired.

8.11.3 RADIUS OF GYRATIONThe radius of gyration (i) is defined by:

i I A ( )= / (8.51)

whereI moment of inertia of the uncracked concrete

section (see Table 2.4)A area of the uncracked concrete section

8.11.4 THRESHOLD SLENDERNESS RATIO, λlim(CL. 5.8.3.1, EC 2)

As noted above, the threshold slenderness value,λlim, is a key element of the design procedure as itprovides a simple and convenient way of determin-ing when to take account of first order effects onlyand when to include second order effects. The valueof λ lim is given by:

λ lim = 20ABC/ n (8.52)

whereA = 1/(1 + 0.2ϕef)(if ϕef is not known,

A = 0.7 may be assumed)B = ( )1 2+ ω (if ω is not known,

B = 1.1 may be used)C = 1.7 − rm (if rm is not known,

C = 0.7 may be used)ϕef effective creep coefficientω mechanical reinforcement

ratio = As fyd /(Ac fcd)As total area of longitudinal reinforcementn relative normal force = NEd/(Ac fcd)rm moment ratio = M01/M02 taken

positive when moments producetension on the same face, otherwisenegative

M01,M02 first order end moments, |M02| ≥ |M01|

8.11.2 EFFECTIVE LENGTHFig. 5.7(f) from EC 2, reproduced as Fig. 8.30,suggests that the effective length of a braced mem-ber (lo) can vary between half and the full height ofthe member depending on the degree of rotationalrestraint at column ends i.e.

l/2 < lo < l

The actual value of effective length can be esti-mated using the following expression

l lk

k

k

ko .

.

. = +

+

+

+

0 5 1

0 451

0 451

1

2

2

(8.50)

wherek1, k2 are the relative flexibilities of rotational

restraints at ends 1 and 2 respectively

in whichk = (θ/M)/(EI/l )θ is the rotation of restraining members for

bending moment M (see Fig. 8.30)EI is the bending stiffness of compression

membersl is the clear height of compression member

between end restraints

Note that in theory, k = 0 for fully rigid rota-tional restraint and k = ∞ for no restraint at all, i.e.pinned support. Since fully rigid restraint is rare inpractice, EC 2 recommends a minimum value of0.1 for k1 and k2.

It is not an easy matter to determine the valuesof k1 and k2 in practice because (a) the guidancein EC 2 is somewhat ambiguous with regard to theeffect of stiffness of columns attached to the columnunder consideration and (b) the effect of crackingon the stiffness of the restraining member. The

l

θ

l/2 < lo < l

Fig. 8.30 Effective length of columns partially restrainedat both ends (based on Fig. 5.7f, EC 2).

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363

numerically equal to the sum of the larger elasticend moment, M02, plus any moment due to geo-metric imperfection, NEd.ei, as follows:

MEd = M02 + NEd.ei (8.53)

where

ei is the geometric imperfection =

θiB0

2

in

which θi is the angle of inclination and canbe taken as 1/200 for isolated braced columnsand l0 is the effective length (clause 5.2(7)).According to clause 6.1(4) the minimumdesign eccentricity, e0, is h /30 but notless than 20 mm where h is the depth ofthe section. Note that in BS 8110 thecorresponding limit is h/20.

Once NEd and MEd have been determined, the areaof longitudinal steel can be calculated by straincompatibility using an iterative procedure as dis-cussed in example 3.19 of this book. However, thisapproach may not be practical for everyday designand therefore The Concrete Centre has produceda series of design charts, similar to those in BS8110:Part 3, which can be used to determine thearea of longitudinal steel. Two typical columndesign charts are shown in Fig. 8.32. Example 8.13illustrates the design procedure involved.

8.11.5.2 Design of columns when λ > λ lim(Cl. 5.8.8, EC 2)

When λ > λlim, critical conditions may occur atthe top, middle or bottom of the column. Thevalues of the design moments at these positionsare, respectively:

(i) M02

(ii) M0Ed + M2

(iii) M01 + 0.5M2

M0Ed is the equivalent first order momentincluding the effect of imperfections at about mid-height of the column and may be taken as M0e asfollows:

M0e = (0.6M02 + 0.4M01) ≥ 0.4M02 (8.54)

where M01 and M02 are the first order end momentsincluding the effect of imperfections acting on thecolumn and M02 is the numerically larger of theelastic end moment acting on the column i.e.|M02| > |M01|.

M2 is the nominal second order moment actingon the column and is given by

M2 = NEd.e2 (8.55)

Design of columns

Fig. 8.31 Column buckling modes.

The value of the creep coefficient, ϕef, can becalculated using the guidance in Cl. 5.8.4 of EC 2.However, in many cases the extra design effortrequired may not be justified and the recommendedvalue of 0.7 for factor A should be used. Factor Bdepends upon the area of longitudinal steel, whichwill be unknown at the design stage. It wouldseem reasonable therefore to use the recommendedvalue of 1.1, at least for the first iteration. Factor Carguably has the largest influence on λlim and it isworthwhile calculating its value accurately ratherthan simply assuming it is equal to 0.7. Factor Cgives an indication of the column’s susceptibilityto buckling under the action of applied moments.Thus buckling is more likely where the end mo-ments act in opposite senses as they will producetension on the same face. Conversely, buckling isless likely when the end moments act in the samesense as the member will be in double curvature(Fig. 8.31).

8.11.5 DESIGN OF BRACED COLUMNSHaving determined the value of λlim it is possibleto design the column. The following sub-sectionsdiscuss the procedures recommended in EC 2 forthe design of braced columns when:

(a) λ < λlim

(b) λ > λlim

8.11.5.1 Design of columns when λ < λ lim(Cl. 6.1, EC 2)

According to clause 5.8.3.1 of EC 2, if the slender-ness, λ, is less than λlim the column should bedesigned for the applied axial action, NEd, andthe moment due to first order effects, MEd, being

9780415467193_C08 11/3/09, 11:14 AM363


364

(a) First order moments for ‘stocky’ columns

(b) Additional second order moments for ‘slender’ columns

(c) Total moment diagram for ‘slender’ columns

M01 + 0.5M20.5M2M01

M02

M2 = NEde2M0e + M2

M0e + M2

e1NEdM

M02

+ =

0

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

N/b

hfck

N/b

hfck

0.05

(a) (b)

0.10 0.15

0.2

0.3

0.41.0

0.90.8

0.5

0.6

0.7

0.8

0.9

Kt = 1

M/bh2fck M/bh2fck

0.20 0.25 0.30 0.35

d2/h = 0.20Asfyk/bhfyk

Asfyk/bhfyk

0.70.6

0.50.4

0.30.2

0.10

0

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00.05 0.10 0.15

0.2

0.3

0.41.0

0.90.8

0.5

0.6

0.7

0.8

0.9

Kt = 1

0.20 0.25 0.30 0.400.35

d2/h = 0.15

0.70.6

0.50.4

0.30.2

0.10

h

bd2

As

2

As

2

h

b

As

2

As

2 d2

Fig. 8.32 Typical column design charts for use with EC 2.

whereNEd is the design axial load at ULS

e2 is the deflection =

1102

r

B0 / (8.56)

in which

1 1r

K Kr

=

r

oϕ (8.57)

where

K

n nn nr

u

u bal

.=

−−

≤ 1 0 (8.58)

in which nu = 1 + ω(where ω = As fyd/Ac fcd) (8.59)

n relative axial force = NEd/Ac fcd (8.60)nbal = 0.4Kϕ = 1 + βϕef ≥ 1.0 (8.61)

in whichϕef effective creep coefficient determined via

clause 5.8.4

β = 0.35 + fck/200 − λ/150 (8.62)

λ slenderness ratio =

Bo

i

and where

10 45r do

yd .

=ε

(8.63)

in which

εyd design yield strain of reinforcement: =

fE

yd

sd effective depth of section

Once NEd and MEd are known, the area of longi-tudinal steel can be evaluated using appropriatecolumn design charts (see Example 8.13).

8.11.6 BIAXIAL BENDING (5.8.9, EC 2)Columns subjected to axial load and bi-axial bend-ing can be checked using the following expression

MM

MM

Edz

Rdz

a

Edy

Rdy

a

+

≤ .1 0 (8.64)

whereMEdz/y design moment around the respective

axis, including a 2nd order momentMRdz/y moment of resistance in the respective

direction

Fig. 8.33

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365

Reinforcement percentages. The area of longi-tudinal reinforcement, As, should lie within thefollowing limits:

greater of

0 10. Nf

Ed

yd

and 0.002 Ac < As < 0.04Ac

wherefyd design yield strength of the reinforcementNEd design axial compression forceAc cross-section of the concrete

Note that the upper limit should be increased to0.08Ac at laps.

8.11.7.2 Transverse reinforcement (links)(Cl. 9.5.3, EC 2)

Size and spacing of links. The diameter of thelinks should not be less than 6 mm or one quarterof the maximum diameter of the longitudinal bar,whichever is the greater.

The spacing of links along the column should notexceed the least of the following three dimensions:

(i) 20 times the minimum diameter of the longi-tudinal bars

(ii) the lesser lateral dimension of the column(iii) 400 mm

Arrangement of links. With regard to thearrangement of links around the longitudinalreinforcement, EC 2 recommends that:

(a) every longitudinal bar placed in a corner shouldbe supported by a link passing around the barand

(b) no bar within a compression zone should befurther than 150 mm from a restrained bar.

Design of columns

Table 8.17 Exponent a for rectangularcross-sections

NEd /NRd 0.1 0.7 1.0

a = 1.0 1.5 2.0

a is the exponent; a = 2 for circular crosssections and for rectangular sections isobtained from Table 8.17 in which NRd

is the design axial resistance of section= Ac fcd + As fyd where Ac is the gross areaof the concrete section and As is the areaof longitudinal steel

For rectangular sections, use of this equationassumes that the ratio of the relative first ordereccentricities, namely ey/h and ez/b, satisfies the fol-lowing relationship

0.2 ≤ (ey /h) ÷ (ez/b) ≤ 5 (8.65)

Unlike BS 8110, the design procedure in EC 2is iterative and involves estimating an area of longi-tudinal steel which is then checked using expres-sion 8.64. To cut down the number of iterations itwould be sensible to use the approach currentlyrecommended in BS 8110 (section 3.13.5.2(ii)) toobtain an initial estimate of the required area oflongitudinal steel, which can be verified using equa-tion 8.64, as shown in example 8.17.

8.11.7 REINFORCEMENT DETAILS FORCOLUMNS

8.11.7.1 Longitudinal reinforcement(Cl. 9.5.2, EC 2)

Diameter of bars. EC 2 recommends that theminimum diameter of longitudinal bars in columnsis 12 mm.

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366

Example 8.13 Column supporting an axial load and uni-axial bending(EC 2)

An internal column in a multi-storey building is subjected to an ultimate axial load (NEd) of 1600 kN and bendingmoment (M) of 60 kNm including effect of imperfections. Design the column cross-section assuming fck = 30 N mm−2,fyk = 500 N mm−2 and ∆cdev = 5 mm.

CROSS-SECTIONSince the design bending moment is relatively small, use equation 8.59 to size the column:

nu = 1 + ω = 1 + A sfyd/A cfcd

Clause 9.5.2 of EC 2 stipulates that the percentage of longitudinal reinforcement, A s, should generally lie within thefollowing limits:

greater of

0 10. Nf

Ed

yd

and 0.002A c < A s < 0.04A c

Assuming that the percentage of reinforcement is equal to, say, 2 per cent gives

A sc = 0.02A c

Substituting this into the above equation gives

1 6 100 85 30 1 5

10 02 500 1 15

0 85 30 1 5

6. ( . / . )

. ( / . )( . / . )

××

= +×A

AAc

c

c

Hence, A c = 62267 mm2.For a square column b = h = √62267 = 250 mm.Therefore a 300 mm square column is suitable.

LONGITUDINAL STEELDesign moment, MEd

Minimum eccentricity, e0 =

h

30

300

3010 20 = = ≥mm mm

Minimum design moment = e0NEd = 20 × 10−3 × 1.6 × 103 = 32 kNm < MHence MEd = M = 60 kNm assuming λ < λlim

Design chartMinimum cover to links for exposure class XC1, cmin,dur = 15 mm (Table 8.9)Assuming diameter of longitudinal bars (Φ) = 25 mm, minimum cover to main steel for bond, cmin,b = 25 mm and thenominal cover, cnom = cmin,b + ∆cdev = 25 + 5 = 30 mm

Assuming diameter of links, (Φ′) = 8 mm,⇒ minimum cover to links = cnom − Φ′ − ∆cdev = 30 − 8 − 5 = 17 mm > cmin,dur = 15 mm OK

Therefore, d2 = 30 + 25/2 = 42.5 mm

dh

2 42 5300

0 142 .

.= =

Round up to 0.15 and use chart no. 1 (Fig. 8.32).

Longitudinal steel area

Nbhf

Ed

ck

.

.=×

× ×=

1 6 10300 300 30

0 5936

Mbh f

Ed

ck2

6

2

60 10300 300 30

0 074

.=

×× ×

=

9780415467193_C08 11/3/09, 11:16 AM366

367

(a) (b)

1402 kN

1412 kN

Beam A Beam B


A fbhf

As yk

ck

s

.=

×× ×

≈500

300 300 300 28 (Fig. 8.32)

⇒ A s = 1512 mm2

Provide 4H25 (1960 mm2)%A sc/A c = 1960/(300 × 300) = 2.2% (acceptable)

LINKSDiameter of links is the greater of:

(i) 6 mm(ii) 1/4Φ = 1/4 × 25 = 6.25 mm

Spacing of links should not exceed the lesser of:

(i) 20Φ = 20 × 25 = 500 mm(ii) least dimension of column = 300 mm(iii) 400 mm

Therefore, provide H8 links at 300 mm centres.

Design of columns

H8 – 300 4H32

Example 8.14 Classification of a column (EC 2)Determine if column GH shown in Fig. 8.34a should be designed for first or second order effects assuming that itresists the design loads and moments in (b). Assume the structure is non-sway and fck = 25 Nmm−2.

Fig. 8.34

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368

Example 8.14 continuedSLENDERNESS RATIO OF COLUMN GH

Effective height

kbeam A = 2(EI/L)A = 2

275 55012 5 10

3

3E

×× ×

= 1.525 × 106 E

kbeam B = 2(EI/L )B = 2

275 55012 7 10

3

3E

×× ×

= 1.089 × 106 E

kcolumn GH = (EI/L)GH = E

275 27512 3 5 10

3

3

.

×× ×

= 1.362 × 105 E

k

kk k

EE EG

column GH

beamA beamB

.

. . . .=

+=

×× + ×

= ≥1 362 10

1 525 10 1 089 100 052 0 1

5

6 6

kH = 0.1 (since column is assumed to be fully fixed at the base)

l l

kk

kko .

.

. = +

+

++

0 5 10 45

10 45

1

1

2

2

l lo .

.. .

.

. . . = × +

+

++

= =0 5 3500 10 1

0 45 0 11

0 10 45 0 1

0 59 2068 mm

(Note that using Table 3.21 of BS 8110, β = 0.75 ⇒ lo = β.lcol = 0.75 × (3500 − 550) = 2212.5 mm)

Radius of gyrationRadius of gyration, i, is given by

i I A ( )= /

= = = = [( / )/( )] / / . bh bh h3 12 12 275 12 79 4 mm

Slenderness ratioSlenderness ratio, λ, is given by

λ

. = = =

lio 2067

79 426

CRITICAL SLENDERNESS RATIO, λ lim

A = 0.7

B = 1.1

C = 1.7 − rm = 1.7 − (M01/M02) = 1.7 − (−29.4/58.8) = 2.2

n = NEd/A cfcd = 1402 × 103/275 × 275 × (0.85/1.5) × 25 = 1.31

λlim / . . . / . .= = × × × =20 20 0 7 1 1 2 2 1 31 29 6ABC n

Since λ < λlim only first order effects need to be considered.

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369

696 kN

716 kN

(a) (b)

Beam A Beam B

55 kNm

27.5 kNm

Example 8.15 Classification of a column (EC 2)Determine if column PQ should be designed for second order effects assuming it resists the design loads and momentshown in Fig. 8.35b. Further assume the structure is non-sway and fck = 25 Nmm−2.

Fig. 8.35

Design of columns

SLENDERNESS RATIO OF COLUMN PQ

Effective height

kbeam A = 2(EI/L)A = 2

275 55012 5 10

3

3E

×× ×

= 1.525 × 106 E

kbeam B = 2(EI/L)B = 2

275 55012 7 10

3

3E

×× ×

= 1.089 × 106 E

kcolumn PQ = (EI/L)PQ = E

275 27512 7 10

3

3

×× ×

= 6.8085 × 104 E

k

kk k

EE EP

column PQ

beamA beamB

.

. . . .=

+=

×× + ×

= ≥6 8085 10

1 525 10 1 089 100 026 0 1

4

6 6

kQ = 0.1 (since column is assumed to be fully fixed at the base)

l l

kk

kko .

.

. = +

+

++

0 5 10 45

10 45

1

1

2

2

l lo .

.. .

.

. . . = × +

+

++

= =0 5 7000 10 1

0 45 0 11

0 10 45 0 1

0 59 4136 mm

(Alternatively, using the coefficients in Table 3.21 of BS 8110, lo = β.lcol = 0.75 × (7000 − 550) = 4837.5 mm)

Radius of gyrationRadius of gyration, i, is given by

i I A ( )= /

= = = = [( )] / / . bh bh h3 12 12 275 12 79 4/ )/( mm

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370


Slenderness ratioSlenderness ratio, λ, is given by

λ

. = = =

lio 4136

79 452

Critical slenderness ratio, λλλλλlim

A = 0.7

B = 1.1

C = 1.7 − rm = 1.7 − (M01/M02) = 1.7 − (−27.5/55) = 2.2

n = NEd/A cfcd = 696 × 103/275 × 275 × 0.567 × 25 = 0.6493

λlim / . . . / . = = × × × =20 20 0 7 1 1 2 2 0 6493 42ABC n

Since λ > λlim column will need to be designed for second order moments.

Example 8.16 Column design (i) λ < λlim; (ii) λ > λlim (EC 2)Design the columns in Examples 8.14 and 8.15. Assume the effective creep coefficient, ϕef, is 0.87 and ∆cdev = 5 mm.

COLUMN GHLongitudinal steelDesign moment, MEd

ei i =

=

×

=θB0

21

2002068

25.2 mm

Minimum eccentricity, e mm mm0 . = = = ≥

h

30

275

309 2 20

Minimum design moment = e0NEd = 20 × 10−3 × 1402 = 28 kNmFirst order end moment, M02 = M + ei.NEd = 58.8 + 5.2 × 10−3 × 1402 = 66.1 kNmHence design moment, MEd = 66.1 kNm

Longitudinal steel areaAssume:diameter of longitudinal steel, Φ = 32 mmdiameter of links, Φ′ = 8 mmminimum cover for durability, cmin,dur = 25 mm > cmin,b = Φ − Φ′(= 32 − 8 = 24 mm)⇒ nominal cover to reinforcement, cnom = c + ∆cdev = 25 + 5 = 30

Therefore, d2 = Φ/2 + Φ′ + cnom = 32/2 + 8 + 30 = 54 mm

dh

2 54275

0 196 .= =

Use graph with d2/h = 0.20 (Fig. 8.32a)

Nbhf

Ed

ck

.

.=×

× ×=

1 402 10275 275 25

0 7426

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371

Mbh f

Ed

ck2

6

2

66 1 10275 275 25

0 127 .

.=

×× ×

=

A fbhf

As yk

ck

s

.=

×× ×

≈500

275 275 250 7 (Fig. 8.32)

⇒ A s = 2647 mm2

Provide 4H32 (3220 mm2)

LinksDiameter of links is the greater of:

(i) 6 mm(ii) 1/4Φ = 1/4 × 32 = 8 mm



Therefore, provide H8 at 275 mm centres.


COLUMN PQFirst order end moments, M01, M02

ei i =

=

×

=θB0

21

2004136

210.34 mm

ei.NEd = 10.34 × 10−3 × 696 = 7.2 kNm

M01 = MQ + ei.NEd = −27.5 + 7.2 = −20.3 kNm

M02 = MP + ei.NEd = 55 + 7.2 = 62.2 kNm

Equivalent first order moment, M0Ed

M0Ed = M0e = (0.6M02 + 0.4M01) ≥ 0.4M02 = 0.4 × 62.2 = 24.9 kNm

= (0.6 × 62.2 + 0.4 × −20.3) = 29.2 kNm

Nominal second order moment, M2

Assumediameter of longitudinal steel (Φ) = 20 mmdiameter of links (Φ′) = 8 mm

Design of columns

H8 – 275

2H32

2H32

Longitudinalbars 4H32

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372

Example 8.16 continuedminimum cover for durability, cmin,dur = 30 mm > cmin,b (= Φ − Φ′ = 20 − 8 = 16 mm)⇒ nominal cover to reinforcement, cnom = c + ∆cdev = 30 + 5 = 35Thus

d = h − ((Φ/2 + Φ′ + cnom)

= 275 − (20/2 + 8 + 35) = 222 mm

10 45

500 1 15 200 100 45 222

2 176 103

5

r do

yd

.

( / . )/ .

. = =×

×= × −ε

β = 0.35 + fck/200 − λ/150 = 0.35 + 25/200 − 52/150 = 0.1283

Kϕ = 1 + βϕef = 1 + 0.1283 × 0.87 = 1.111 ≥ 1.0

Assume K r = 0.8

1 10 8 1 111 2 176 10 1 934 105 5

rK K

r . . . . =

= × × × = ×− −r

oϕ

e

r22 5 21

10 1 934 10 4136 10 / . / =

= × × =−B0 33.1 mm

M2 = NEd.e2 = 696 × 33.1 × 10−3 = 23 kNm

Design moment, MEd

MEd = maximum of {M0Ed + M2; M02; M01 + 0.5M2}

MEd = maximum of {29.2 + 23 = 52.2 kNm; 62.2 kNm; −20.3 + 0.5 × −23 = 31.8 kNm}

Longitudinal steel area

d2 = Φ/2 + Φ′ + cnom = 20/2 + 8 + 35 = 53 mm

dh

2 53275

0 193 .= =

Use graph with d2/h = 0.2 (Fig. 8.32)

Nbhf

Ed

ck

.=

×× ×

=696 10

275 275 250 368

3

Mbh f

Ed

ck2

6

2

62 2 10275 275 25

0 12 .

.=

×× ×

=

A fbhf

As yk

ck

s

.=

×× ×

=500

275 275 250 3

⇒ A s = 1134 mm2

Provide 4H20 (1260 mm2)

Check assumed value of Kr

nu = 1 + A sfyd/A cfcd = 1 + 1260 × (0.87 × 500)/(275 × 275 × 0.567 × 25) = 1.51

n = NEd/A cfcd = 696 × 103/(275 × 275 × (0.85/1.5) × 25) = 0.65

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373

H6 – 275 4H20

K

n nn nr

u

u bal

. .. .

.=−

−=

−−

=1 51 0 651 51 0 4

0 775

Therefore assumed value is acceptable.

LinksDiameter of links is the greater of:

(i) 6 mm(ii) 1/4Φ = 1/4 × 20 = 5 mm



Therefore, provide H6 links at 275 mm centres.


RELATIVE ECCENTRICITIESThe column is subject to an axial load, NEd = 1250 kN, a moment about the major axis (y-y), MEdy = 35 kNm and amoment about the minor axis (z-z), MEdz = 25 kNm

e

M

Ny

Edy

Ed

mm

= =

××

=35 10

1250 1028

6

3

e

M

Nz

Edz

Ed

mm

= =

××

=25 10

1250 1020

6

3

(e y/h) ÷ (e z/b) = (28/275) ÷ (20/275) = 1.4 > 0.2

Therefore check column for bi-axial bending

Design of columns

Example 8.17 Column subjected to combined axial load and biaxialbending (EC 2)

Check the column in Example 3.22(b) using the procedure in EC 2.

2H25

H8-275

2H25

275 × 275

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374

Example 8.17 continuedDESIGN ULTIMATE MOMENT OF RESISTANCE, MRd

d2 = 35 + 8 + 25/2 = 55.5 mm and d2/h = 55.5/275 = 0.2

A fbhf

s yk

ck

.=

×× ×

=1960 500

275 275 250 518

Nbhf

Ed

ck

.=

×× ×

=1250 10

275 275 250 66

3

Mbh f

Rd

ck2

0 105 .≈ (Fig. 8.32)

⇒ MRd = 54.6 kNm

EXPONENT, aNRd = A cfcd + A sfyd

= [2752 × (0.85 × 25/1.5) + 1960 × (500/1.15)] × 10−3

= 1071.35 + 852.17 = 1923.5 kN

NN

Ed

Rd

.

.= =1250

1923 50 65

From Table 8.17, a ≈ 1.46

RESISTANCE CHECK

MM

MM

Edz

Rdz

aEdy

Rdy

a 1.46 1.46

+

≤ =

××

+××

= + = <

.

.

. . . 125 10

54 6 1035 10

54 6 100 32 0 52 0 84 1

6

6

6

6

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375

Chapter 9

Eurocode 3: Designof steel structures

This chapter describes the contents of selected sub-partsof Part 1 of Eurocode 3, the new European standardfor the design of buildings in steel, which is expected toreplace BS 5950 by about 2010. The chapter highlightsthe principal differences between Eurocode 3 and BS5950 and illustrates the new design procedures by meansof a number of worked examples on beams, columnsand connections. To help comparison but primarily toease understanding of the Eurocode, the material hashere been presented in a similar order to that in Chap-ter 4 of this book on BS 5950.

9.1 IntroductionEurocode 3 applies to the design of buildings andcivil engineering works in steel. It is based on limitstate principles and comes in several parts as shownin Table 9.1.

Part 1, which is the subject of this discussion,gives the basic rules for design of buildings in steel.It is largely similar in scope to BS 5950: Part 1,which it will replace by about 2010. Originallyplanned as a three sub-part document, the norm-ative version was eventually published in twelvesub-parts as indicated in Table 9.2. This step wastaken in order to avoid repetition of design guidanceand provide a concise set of standards for design ofsteel buildings. But this also means that engineers

Table 9.1 Overall scope of Eurocode 3

Part Subject

1 General rules and rules for buildings2 Steel bridges3 Towers, masts and chimneys4 Silos, tanks and pipelines5 Piling6 Crane supporting structures

Table 9.2 Scope of Part 1 of Eurocode 3

Part Subject

1.1 General rules and rules for buildings1.2 Structural fire design1.3 Cold formed thin gauge members and sheeting1.4 Stainless steels1.5 Plated structural elements1.6 Strength and stability of shell structures1.7 Strength and stability of planar plated structures

transversely loaded1.8 Design of joints1.9 Fatigue strength of steel structures1.10 Selection of steel for fracture toughness and

through thickness properties1.11 Design of structures with tension components

made of steel1.12 Supplementary rules for high strength steel

will now have to refer to a number of standardsin order to design even the simplest of elements.In the case of beams, for example, the designerwill have to refer to Part 1.1, hereafter referred toas EC 3, for guidance on bending and shear, Part1.5, hereafter referred to as EC 3–5, for web andstiffener design and Part 1.8, hereafter referred toas EC 3–8, for connection design. In addition, butlike the other material-specific codes, i.e. EC 2–EC6 and EC 9, the designer will have to refer to EN1990 and Eurocode 1 to determine the designvalues of actions and combination of actions, includ-ing values of the partial factors for actions.

EC 3 was published in draft form in November1990 and then as a voluntary standard, DD ENV1993–1–1, in September 1992. It was finallyissued as a Euronorm, EN 1993–1–1, but with amuch reduced content, in 2004. EC 3 deals withthe following subjects:

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Eurocode 3: Design of steel structures

376

Chapter 1: GeneralChapter 2: Basis of designChapter 3: MaterialsChapter 4: DurabilityChapter 5: Structural analysisChapter 6: Ultimate limit stateChapter 7: Serviceability limit stateAnnex A – Method 1: Interaction factorsAnnex B – Method 2: Interaction factorsAnnex AB – Additional design provisionsAnnex BB – Buckling of components of building

structures

The purpose of the following discussion is prin-cipally to highlight the main differences betweenEC 3, EC 3–5, EC 3–8 and BS 5950: Part 1.A number of comparative design studies are alsoincluded to illustrate the new design procedures.

9.2 Structure of EC 3Apart from the major re-packaging exercise referredto above, the organisation of material within indi-vidual sub-parts is also quite different to that usedin BS 5950, as exemplified by the above contentslist for EC 3. Thus, whereas BS 5950 containsseparate sections on the design of different types ofelements in building structures, e.g. beams, plategirders, compression members, etc., EC 3, on theother hand, divides the material on the basis ofstructural phenomena, e.g. tension, compression,bending, shear and buckling, which may apply toany element. This change has largely been imple-mented in order to avoid duplication of designrules and also to promote a better understandingof structural behaviour.

9.3 Principles and Applicationrules (Cl. 1.4, EC 3)

As with the other structural Eurocodes and for thereasons discussed in section 7.5.2, the clauses inEC 3 have been divided into Principles and Applica-tion rules. Principles comprise general statements,definitions, requirements and models for which noalternative is permitted. Principles are indicatedby the letter P after the clause number. The Appli-cation rules are generally recognised rules whichfollow the statements and satisfy the requirementsgiven in the Principles. The absence of the letterP after the clause number indicates an applicationrule. The use of alternative application rules tothose recommended in the Eurocode is permitted

provided it can be shown that the alternatives areat least equivalent and do not adversely affect otherdesign requirements. It is worth noting, however,that if an alternative application rule is used theresulting design will not be deemed Eurocodecompliant (Fig. 9.1).

9.4 Nationally DeterminedParameters

Like the other structural Eurocodes, Eurocode 3allows some parameters and design methods to bedetermined at the national level. Where a nationalchoice is allowed this is indicated in a note in thenormative text under the relevant clause. The notemay include the recommended value of the par-ameter, or preferred method, etc., but the actualvalue, methodology, etc., to be used in a particularMember State country is/will be given in the appro-priate National Annex. The recommended valuesof these parameters and design methods/proceduresare collectively referred to as Nationally DeterminedParameters (NDPs). The NDPs determine variousaspects of design but perhaps most importantlythe level of safety of structures during execution(i.e. construction/fabrication) and in-service, whichremains the responsibility of individual nations.

The UK National Annexes to EC 3 and EC 3–5were published in 2008 and 2005 respectively. TheNational Annex to EC 3–8 is still awaited. Thusthe discussion and worked examples in sections 9.10and 9.11 on, respectively, beam and column design,are based on the material in EC 3, EC 3–5 andthe NDPs in the accompanying National Annexes

6.2.4 Compression

(1)P The design value of the compression force NEd ateach cross-section shall satisfy:

NN

Ed

c,Rd

,≤ 10 (6.9)

(2) The design resistance of the cross-section foruniform compression Nc,Rd should be determined asfollows:

NAf

c,Rdy

M

= γ 0

for class 1, 2, or 3 cross-sections (6.10)

NA f

c,Rdeff y

M

= γ 0

for class 4 cross-sections (6.11)

Fig. 9.1 Example of Principle and Application rules (Cl.6.2.4, EC 3, p49).

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377

Basis of design

whereas the material in section 9.12 on connec-tions is based solely on the contents of EC 3–8.

9.5 Symbols (Cl. 1.6, EC 3)The symbols used in EC 3 are extremely schematicbut a little cumbersome. For example, Mpl,y,Rd

denotes the design plastic moment of resistanceabout the y–y (major) axis. Symbols such as thismake many expressions and formulae much longerand seem more complex than they actually are.

A number of symbols which are used to identifyparticular dimensions of Universal sections havechanged in Eurocode 3 as discussed in section 9.6.For example, the elastic and plastic moduli aboutthe major axis are denoted in Eurocode 3 by thesymbols Wel,y and Wpl,y respectively rather than Zxx

and Sxx as in BS 5950. Likewise, the elastic andplastic moduli about the minor axis are denoted inEC 3 by the symbols Wel,z and Wpl,z, respectively,rather than Zyy and Syy as in BS 5950.

9.6 Member axes (Cl. 1.6.7, EC 3)Member axes for commonly used steel membersare shown in Fig. 9.2. This system is different tothat used in BS 5950. Thus, x–x is the axis alongthe member, while axis y–y is what is termed inBS 5950 as axis x–x, the major axis. The minor axisof bending is taken as axis z–z in EC 3. Considering

steel design in isolation, this change need not havehappened. It was insisted on for consistency withthe structural Eurocodes for other materials, e.g.concrete and timber.

9.7 Basis of designLike BS 5950, EC 3 is based on the limit statemethod and for design purposes principally con-siders two categories of limit states: ultimate andserviceability. A separate (third) category of dura-bility is also mentioned in clause 4 of EC 3 whichcovers the limit states of corrosion, mechanical wearand fatigue. Of these, corrosion is generally viewedas the most critical issue for steel building struc-tures and involves taking into account the expectedexposure conditions, the design working life of thestructure, the composition and properties of thesteel, structural detailing and protective measures,amongst other factors (clause 2.4, EN 1990).

The terms ‘ultimate limit state’ (ULS) and ‘ser-viceability limit state’ (SLS) apply in the same wayas we understand them in BS 5950. Thus, ultimatelimit states are those associated with collapse, orwith other forms of structural failure which mayendanger the safety of people while serviceabilitylimit states concern states beyond which specifiedservice criteria, for example the functioning of thestructure or member, the comfort of people andappearance of the structure, are no longer met(clauses 3.3 and 3.4 of EN 1990).

Fig. 9.2 Member definitions and axes used in EC 3 (NB symbols in parentheses denote symbols used in BS 5950).

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378

The actual serviceability limit states relevant tosteel building structures are outlined in section 7of EC 3. The code provides very few details, how-ever, as many SLS are not material specific and forrelevant design guidance the designer must turn toclause A1.4 of EN 1990. The principal serviceabilitylimit states recommended in EC 3 are listed below:

1. Deformations, e.g. vertical and horizontaldeflections which cause damage to finishes ornon-structural elements such as partitions andcladding, or adversely affect the comfort ofpeople or appearance of the structure.

2. Dynamic effects, e.g. vibration, oscillation orsway which causes discomfort to the occupantsof a building or damage to its contents.

Based on the recommendations given in clauses3.2 and 3.3 of EN 1990, the ultimate limit statesrelevant to steel structures and components are:

(1) static equilibrium of the structure(2) rupture, excessive deformation of a member,

transformation of the structure into amechanism

(3) instability induced by second order effects, e.g.lack of fit, thermal effects, sway

(4) fatigue(5) accidental damage, e.g. exposure to fire, explo-

sion and impact.

As can be appreciated, these are roughly similarto the ultimate limit states used in BS 5950(Table 4.1) although there are slight changes in theterminology.

In the context of elemental design, the designeris principally concerned with the ultimate limit stateof rupture (i.e. strength) and the serviceability limitstate of deflection. EC 3 requirements in respect ofthese two limit states will be discussed later forindividual element types.

In a similar way to BS 5950, clauses 1.5.3 and5.1 of EC 3 distinguishes between three types offrames used for building structures and associatedjoint behaviour:

(a) simple frames in which the joint are notrequired to resist moments, i.e. the joints arepinned and by implication the structure is fullybraced.

(b) continuous frames in which the joints are rigid(for elastic analysis) or full strength (for plasticanalysis) and can transmit all moments andforces.

(c) semi-continuous frames in which the actualstrength of the joints is taken into account.

9.8 Actions (Cl. 2.3, EC 3)Actions is Eurocode terminology for loads andimposed deformations. Permanent actions, G, areall the fixed loads acting on the structure, includingfinishes, immovable partitions and the self weightof the structure. Variable actions, Q, include theimposed, wind and snow loads. Clause 2.3.1 ofEC 3 recommends that the values of characteristicpermanent, Gk, and variable, Qk, actions should betaken from relevant parts of Eurocode 1: Actions onstructures. Guidance on determining the design valueof actions and combination of actions is given inEN 1990: Basis of structural design. These documentsand topics are briefly discussed in section 8.5 ofthis book. As noted there, the design value of anaction (Fd) is obtained by multiplying the repre-sentative value (Frep) by the appropriate partial safetyfactor for actions (γ f):

Fd = γ f Frep (9.1)

Table 9.3 shows the relevant partial safety factorsfor the ultimate limit state of strength. Other safetyfactors will apply in other design situations. Forexample, the partial factors for the ultimate limitstates of equilibrium are shown in Table 8.6. Inequation 9.1, Frep is generally taken as the charac-teristic value of a permanent or variable action (i.e.Fk). Assuming that the member being designed issubjected to one or more permanent actions andone variable action only, i.e. load combination 1 inTable 9.3, the partial safety factor for permanentactions, γG can conservatively be taken as 1.35 andfor the variable action, γQ as 1.5. The correspond-ing values of the safety factors in BS 5950 are 1.4and 1.6 respectively. As discussed in section 8.5.3,it is possible to improve structural efficiency byusing expressions 6.10a and 6.10b of EN 1990(respectively load combination 3(a) and 3(b)/3(c)in Table 9.3) to estimate the design values of actionsbut the value of 1.35 for γG is conservative andused throughout this chapter.

9.9 Materials

9.9.1 DESIGN STRENGTHS (CL. 2.4.3, EC 3)The design strengths, Xd, are obtained by dividingthe characteristic strengths, Xk, which can be takenas nominal values in calculations, by the materialpartial safety factor, γm, i.e.

X

Xd

k

m

=γ

(9.2)

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379

Materials





Strength:1. Permanent and variable 1.35/1.35ξ 1.0 1.5 0 –2. Permanent and wind 1.35/1.35ξ 1.0 – – 1.53. Permanent, imposed and wind

(a) 1.35 1.0 1.5ψ0,1 0 1.5ψ0,2

(b) 1.35/1.35ξ 1.0 1.5 0 1.5ψ0

(c) 1.35/1.35ξ 1.0 1.5ψ0 0 1.5

But this is in fact not used in EC 3. Instead,EC 3 divides the cross-section resistance, Rk, by apartial safety factor to give the design resistance,Rd for the member, i.e.

R

Rd

k

M

=γ

(9.3)

where γM is the partial safety factor for theresistance. Thus, γM in EC 3 is not the same as γm

in BS 5950. Here γm is applied to material strength;γM is applied to members or structures and takesaccount of both material and modelling/geometricuncertainties.

9.9.2 NOMINAL STRENGTHS (CL. 3.2, EC 3)Table 9.4 shows the steel grades and associatednominal values of yield strength and ultimate

tensile strength for hot rolled steel sections. The steelgrades are the same as those mentioned in BS 5950except grade S235 which is popular in Continentalcountries but not in the UK and was thereforenot referred to in BS 5950. Note that the values ofthe yield strength shown in the table have beenobtained from the product standard (EN 10025–2)and not from EC 3, as recommended in clause 2.4of the National Annex.

9.9.3 PARTIAL FACTORS, γM (CL. 6.1, EC 3)Values of the partial factor γM (section 9.9.1)applied to characteristic values of resistance toobtain design resistances are given in clause 6.1of EC 3. The factor γM assumes different valuesdepending on the type of resistance being verifiedas indicated below:

Table 9.4 Nominal values of yield strength fy and ultimate tensilestrength fu for hot rolled structural steel to EN 10025–2 (Table 3.1,EC 3)

Nominal Nominal thickness of the element, t (mm)steelgrade t ≤ 16 mm 16 mm < t ≤ 40 mm

fy(N/mm2) fu(N/mm2) fy(N/mm2) fu(N/mm2)

S235 235 360 235 360S275 275 410 265 410S355 355 470 345 470S450 440 550 430 550

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380

resistance of cross-sections γM0 = 1.00resistance of member to instability γM1 = 1.0resistance of cross-section to fracture γM2 = 1.10

The γM factors for joints are given in EC 3–8 asdiscussed in Section 9.13.

9.9.4 MATERIAL COEFFICIENTS (CL. 3.2.5, EC 3)The following coefficients are specified in EC 3

Modulus of elasticity, E = 210,000 N mm−2

Shear Modulus,

G

F

( ) =

+≈ −

2 181000 2

νN mm (9.4)

Poisson’s ratio, ν = 0.3Coefficient of linear thermal expansion, α = 12 ×

106 per deg CDensity, ρ = 7850 kg m−3

Note that the value of E is slightly higher thanthat specified in BS 5950 (205 kN mm−2).

9.10 Classification ofcross-sections (Cl. 5.5, EC 3)

Classification has the same purpose as in BS 5950,and the four classifications are identical:

Class 1 cross-sections: ‘plastic’ in BS 5950Class 2 cross-sections: ‘compact’ in BS 5950Class 3 cross-sections: ‘semi-compact’ in BS 5950Class 4 cross-sections: ‘slender’ in BS 5950

Classification of a cross-section depends uponthe proportions of each of its compression elements.The highest (least favourable) class number shouldbe quoted for a particular section.

Appropriate sections of EC 3’s classificationare given in Table 9.5. Comparison with Table 5.2of EC 3 shows that the symbols are slightly differ-ent. This change has been carried to prevent anyconfusion as to which dimension is to be used toclassify sections as EC 3 uses the same symbols todefine different section dimensions. Comparisonwith the corresponding limits given in BS 5950for rolled sections (Table 4.4) shows that the c*/tw

(= d/t) ratios are slightly more onerous for webs,and c/t f (= (B − tw − 2r)/2T ) ratios are perhapsslightly less onerous for outstand flanges. It shouldbe noted that in EC 3 the factor ε is given by

ε = (235/fy)0.5 (9.5)

not (275/py)0.5 as in BS 5950.

For Class 4 sections effective cross-sectionalproperties can be calculated using effective widths

Table 9.5 Maximum width-to-thickness ratiosfor compression elements (Table 5.2, EC3)

Type of element Class of element

Class 1 Class 2 Class 3

Outstand flange

ct f

≤ 9ε

ct f

≤ 10ε

ct f

≤ 14εfor rolledsection

Web with

ct*

w

≤ 72ε

ct*

w

≤ 83ε

ct*

w

≤ 124εneutral axis atmid depth,rolled section

Web subject to

ct*

w

≤ 33ε

ct*

w

≤ 38ε

ct*

w

≤ 42εcompression,rolled sections

fy 235 275 355ε 1 0.92 0.81

c* = d; c = (b − tw − 2r)/2

of plate which in some cases put the memberback into Class 3. The expressions for calculatingeffective widths of hot rolled sections are given inEC 3–5.

9.11 Design of beamsAs noted at the outset, the information neededfor beam design is no longer in one place butdispersed in EC 3, EC 3–5 and EN 1990. To easeunderstanding of this material, as in Chapter 4 ofthis book, we will consider the design of fully later-ally restrained and unrestrained beams separately.Thus, the following section (9.11.1) will consider thedesign of beams which are fully laterally restrained.Section 9.11.2 will then look at Eurocode 3: Part 1’srules for designing beams which are not laterallytorsionally restrained.

9.11.1 FULLY LATERALLY RESTRAINEDBEAMS

Generally, such members should be checked for

(1) resistance of cross-section to bending ULS(2) resistance to shear buckling ULS(3) resistance to flange-induced buckling ULS(4) resistance of the web to transverse forces ULS(5) deflection SLS.

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Design of beams

9.11.1.1 Resistance of cross-sections-bendingmoment (Cl. 6.2.5, EC 3)

When shear force is absent or of a low value, thedesign value of the bending moment, MEd, at eachsection should satisfy the following:

MM

Ed

c,Rd

.≤ 1 0 (9.5)

where Mc,Rd is the design resistance for bendingabout one principal axis taken as follows:

(a) the design plastic resistance moment of thegross section

Mpl,Rd = Wp1 fy /γM0 (9.6)

where Wpl is the plastic section modulus, forclass 1 and 2 sections only;

(b) the design elastic resistance moment of the grosssection

Mel,Rd = Wel,min fy /γM0 (9.7)

where Wel,min is the minimum elastic sectionmodulus for class 3 sections;

(c) the design local buckling resistance momentof the gross section

Mc,Rd = Weff,min fy /γM0 (9.8)

where Weff,min is the minimum effective sectionmodulus, for class 4 cross-sections only;

(d) the design ultimate resistance moment of the netsection at bolt holes Mu,Rd, if this is less thanthe appropriate values above. In calculating thisvalue, fastener holes in the compression zonedo not need to be considered unless they areoversize or slotted and provided that they arefilled by fasteners. In the tension zone holes donot need to be considered provided that

A f A ff,net u

M2

f y

M0

0 9.

γ γ≥ (9.9)

Consideration of fastener holes in bending is adifference from BS 5950.

9.11.1.2 Resistance of cross-sections – shear(Cl. 6.2.6, EC 3)

The design value of the shear force VEd at eachcross-section should satisfy the following

VV

Ed

c,Rd

.≤ 1 0 (9.10)

where Vc,Rd is the design shear resistance. Forplastic design Vc,Rd is taken as the design plasticshear resistance, Vpl,Rd, given by

Vpl,Rd = Av( fy / 3 )/γM0 (9.11)

where Av is the shear area, which for rolled I andH sections, loaded parallel to the web is

Av = A − 2bt f + (tw + 2r)t f ≥ ηhwt f (9.12)

whereA cross-sectional areab overall breadthr root radiustf flange thicknesstw web thicknesshw depth of the webη conservatively taken as 1.0

Equation 9.12 will generally give a slightly higherestimate of the shear area than in BS 5950.

Fastener holes in the web do not have to beconsidered in the shear verification.

Shear buckling resistance for unstiffened websmust additionally be considered when

ht

w

w

> 72εη

(9.13)

When the changed definition of ε is taken intoaccount, this value is more onerous than the corres-ponding value in BS 5950.

For a stiffened web (clause 5.1, EC 3–5), shearbuckling resistance will need to be considered when

ht

kw

w

>31η

ε τ (9.14)

where kτ is the buckling factor for shear and isgiven by

for a/hw < 1 (Fig. 9.3)

kτ = 4 + 5.34(hw/a)2 (9.15)

for a/hw ≥ 1kτ = 5.34 + 4(hw/a)2 (9.16)

9.11.1.3 Resistance of cross-sections-bendingand shear (Cl. 6.2.8, EC 3)

The plastic resistance moment of the section isreduced by the presence of shear. When the designvalue of the shear force, VEd, exceeds 50 per cent ofthe design plastic shear resistance, Vpl,Rd, the designresistance moment of the section, Mv,Rd, should becalculated using a reduced yield strength taken as

(1 − ρ)fy for the shear area only (9.17)

where

ρ = −

2

1

2V

VEd

pl,Rd

(9.18)

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382

FS FS FS

V1,S V2,S VShw

sS sS sSc

a

Type (a) Type (b) Type (c)

Fig. 9.3 Different types of forces applied to a web and associated buckling coefficients (Fig. 6.1, EC 3–5).

khaFw = +

6 22

khaFw . = +

3 5 22

ks c

hFs

w

= ++

≤2 6 6

Thus, for rolled I and H sections the reduceddesign resistance moment for the section about themajor axis, My,v,Rd, will be given by

My,v,Rd = fy(Wpl,y − ρAv2/4tw)/γM0 ≤ My,c,Rd (9.19)

9.11.1.4 Shear buckling resistance(Cl. 6.2.6, EC 3)

As noted above the shear buckling resistance ofunstiffened beam webs has to be checked when

ht

w

w

> 72εη

(9.20)

Clause 2.4 of the National Annex to EC 3–5recommends a value of 1.0 for η for all steel gradesup to and including S460. For standard rolledbeams and columns this check is rarely necessary,however, as hw/tw is almost always less than 72ε andhas therefore not been discussed in this section.

9.11.1.5 Flange-induced buckling (Cl. 8, EC 3–5)To prevent the possibility of the compression flangebuckling in the plane of the web, EC 3–5 requiresthat the ratio hw/tw of the web should satisfy thefollowing criterion:

ht

kEf

AA

w

w yf

w

fc

≤ (9.21)

whereAw is the area of the web = (h − 2t f)tw

Afc is the area of the compression flange = btf

fyf is the yield strength of the compression flange

The factor k assumes the following values:

Plastic rotation utilised, i.e. class 1 flanges: 0.3Plastic moment resistance utilised, i.e. class 2flanges: 0.4Elastic moment resistance utilised, i.e. class 3 orclass 4 flanges: 0.55

9.11.1.6 Resistance of the web to transverseforces (Cl. 6, EC 3–5)

EC 3–5 distinguishes between two types of forcesapplied through a flange to the web:

(a) forces resisted by shear in the web (Fig. 9.3,loading types (a) and (c)).

(b) forces transferred through the web directly tothe other flange (Fig. 9.3, loading type (b)).

For loading types (a) and (c) the web is likely tofail as a result of

(i) crushing of the web close to the flange accom-panied by yielding of the flange, the combinedeffect sometimes referred to as web crushing

(ii) localised buckling and crushing of the webbeneath the flange, the combined effect some-times referred to as web crippling (Fig. 9.4a).

For loading type (b) the web is likely to fail as aresult of

(i) web crushing(ii) buckling of the web over most of the depth of

the member (Fig. 9.4b).

Provided that the compression flange is adequatelyrestrained in the lateral direction, the design resist-ance of webs of rolled beams under transverse forcescan be determined in accordance with the recom-mendations contained in clause 6 of EC 3–5.

Here it is stated that the design resistance ofwebs to local buckling is given by

F

f L tRd

yw eff w

M1

=γ

(9.22)

wherefyw is the yield strength of the webtw is the thickness of the webγM1 is the partial safety factor = 1.0Leff is the effective length of web which resists

transverse forces = χF ly (9.23)

9780415467193_C09 9/3/09, 4:20 PM382

383

FS

sS

45°

Design of beams

Loading types (a) and (c) (Fig 9.3) Loading type (b) (Fig 9.3)

Fig. 9.4 Web failure.

in whichχF is the reduction factor due to local buckling

calculated as discussed belowly is the effective loaded length, appropriate to

the length of the stiff bearing ss. Accordingto clause 6.3 of EC 3–5, ss should be takenas the distance over which the applied loadis effectively distributed at a slope of 1.1 asshown in Fig. 9.5, but ss ≤ h w.

Reduction factor, χF

According to clause 6.4 the reduction factor χF isgiven by

χF

F

.

= ≤0 5

1d

(9.24)

where

d

BF

y w yw

cr

=t fF

(9.25)

in which

F k E

thcr F

w3

w

.= 0 9 (9.26)

For webs without longitudinal stiffeners kF

is obtained from Fig. 9.3 and ly is obtained asfollows.

Effective loaded length, lyAccording to clause 6.5 for loading types (a) and(b) in Fig. 9.3, the effective loaded length, ly, isgiven by

By = ss + 2t f 1 + +( )m m1 2 ≤ a (9.27)

where

m

f bf t1

yf f

yw w

= (9.28)

and if dF > 0.5

m

ht2

w

f

.=

0 022

(9.29)

or if dF ≤ 0.5, m2 = 0For loading type (c) ly is taken as the smallest

value obtained from equations 9.30 and 9.31 asfollows

ly e f

1 e

f

= + +

+BB

tm

tm

2

2

2(9.30)

B By e f 1 = + +t m m2 (9.31)

Fig. 9.5 Length of stiff bearing (based on Fig. 6.2,EC 3–5).

9780415467193_C09 9/3/09, 4:20 PM383


384

where

Be

F w2

yw ws= ≤ +

k Etf h

s c2

(9.32)

Interaction between shear force and bendingmomentAccording to clause 7 of EC 3–5, where the web isalso subject to bending the combined effect shouldsatisfy the following

η2 + 0.8η1 ≤ 1.4 (9.33)

where

η

γ

1 1 0 .= ≤M

f WEd

y pl

M0

(9.34)

η

γ

2 1 0 .= ≤F

f L tEd

yw eff w

M1

(9.35)

9.11.1.7 Deflections (Cl. 7, EC 3)Clause 7 of EC 3 highlights the need to check theSLS of deflection but provides few details on thesubject other than to refer the designer to Annex

A1.4 of EN 1990. Clause 7.2.1 of EC 3 makesclear, however, that the designer is responsible forspecifying appropriate limits for vertical deflections,which should be agreed with the client.

Several vertical deflections are defined in EN1990. However, like BS 5950, the National Annexto EC 3 recommends that checks on the verticaldeflections, δ, under unfactored imposed loadingshould be carried out and suggests that in theabsence of other limits the limits shown in Table9.6 may be used.

DESIGN BENDING MOMENT

Design action (FEd) = (γGgk + γQqk) × span

= (1.35 × 8 + 1.5 × 6)8 = 158.4 kN

Design bending moment (MEd) =

F lEd kNm8

158 4 88

158 4 .

. =×

=

STRENGTH CLASSIFICATIONFlange thickness, tf = 11.5 mm, steel grade S275. Hence from Table 9.4, fy = 275 N mm−2

Example 9.1 Analysis of a laterally restrained beam (EC 3)Check the suitability of 356 × 171 × 51 kg m−1 UB section in S275 steel loaded by uniformly distributed loadinggk = 8 kN m−1 and qk = 6 kN m−1 as shown below. Assume that the beam is fully laterally restrained and that thebeam sits on 100 mm bearings at each end. Ignore self weight of beam.

Table 9.6 Recommended vertical deflectionlimits (Clause 2.24 of National Annex to EC 3)

Conditions Deflection limits

Cantilevers Length/180Beams carrying plaster Span/360

or other brittle finishOther beams Span/200

(except purlinsand sheeting rails)

Purlins and sheeting rails To suit the characteristicsof particular cladding

9780415467193_C09 9/3/09, 4:20 PM384

385

Design of beams

SECTION CLASSIFICATION

ε = (235/fy)0.5 = (235/275)0.5 = 0.92

ct f

..

. . .= = < = × =71 911 5

6 25 9 9 0 92 8 28ε

(where c = (b − tw − 2r)/2 = (171.5 − 7.3 − 2 × 10.2)/2 = 143.8/2 = 71.9 mm)

ct*

w

.

. . . .= = < = × =

312 37 3

42 8 72 72 0 92 66 2ε

Hence from Table 9.5, 356 × 171 × 51 UB belongs to Class 1.

RESISTANCE OF CROSS-SECTION

Bending momentSince beam section belongs to class 1, plastic moment of resistance, Mpl,Rd, is given by

M

W fpl,Rd

pl,y y

M0

.= =

× ×γ

895 10 2751 00

3

= 246.1 × 106 Nmm

= 246.1 kNm > MEd (= 158.4 kNm) OK

ShearDesign shear force, VEd, is

V

FEd

Ed kN .

. = = =2

158 4

279 2

For class 1 section design plastic shear resistance, Vpl,Rd, is given by

Vpl,Rd = A v(fy/ 3 )/γM0

where A v = A − 2btf + (tw + 2r )t f ≥ ηhwtw = 1.0 × (355.6 − 2 × 11.5) × 7.3 = 2428 mm2

= 64.6 × 102 − 2 × 171.5 × 11.5 + (7.3 + 2 × 10.2)11.5

= 2834 mm2 OK

Hence, Vpl,Rd = 2834(275/ 3 )/1.00

= 540371 N = 540.4 kN > VEd (= 79.2 kN) OK

Bending and shearAccording to EC 3 the theoretical plastic resistance moment of the section, i.e. Mpl,Rd, is reduced if

VEd > 0.5Vpl,Rd

But in this case

VEd = 0 (at mid-span)

Hence, no check is required.

SHEAR BUCKLING RESISTANCE

As

ht

w

w

.

. .

..

.= = < = × =332 67 3

45 6 72 720 921 0

66 24εη


9780415467193_C09 9/3/09, 4:21 PM385


386

Example 9.1 continuedwhere

hw = h − 2tf = 355.6 − 2 × 11.5 = 332.6 mm

No check on shear buckling is required.

FLANGE-INDUCED BUCKLINGAw, area of web = (h − 2tf)tw = (355.6 − 2 × 11.5) × 7.3 = 2428 mm2

Afc, area of compression flange = btf = 171.5 × 11.5 = 1972.2 mm2

ht

kEf

AA

w

w yf

w

fc

. .

. .= ≤ = ×

×=45 6 0 3

210 10275

24281972 2

254 23

OK


WEB BUCKLINGAt an unstiffened end support (i.e. load type (c) in Fig. 9.3)

k

s chF

s

w

= ++

≤2 6 6

= +

+

= .

.2 6100 0

332 63 80

F k E

t

hcr F

w

w

. . . .

. N= = × × × × =0 9 0 9 3 80 210 10

7 3

332 6840024

33

3

m

f bf t1

yf f

yw w

. .

.= =××

=275 171 5275 7 3

23 5

Assuming d > 0.5

m

ht2

w

f

. . .

. .=

= ×

=0 02 0 02332 611 5

16 732 2

Be

F w2

yw ws mm = ≤ + = + =

k Etf h

s c2

100 0 100

=

× × ×× ×

= . .

. .

3 80 210 10 7 3

2 275 332 6232 5

3 3

mm

Hence le = 100 mm

B B

By e f

1 e

f

= + +

+tm

tm

2

2

2

= + +

+ = ..

.

. 100 11 523 5

210011 5

16 72

217.3 mm

B By e f 1 = + +t m m2

By = 100 + 11 5 23 5 16 7. . .+ = 172.9 m

9780415467193_C09 9/3/09, 4:21 PM386

387

Design of beams

Example 9.1 continuedHence, ly = 172.9 mm

d

BF

y w yw

cr

. .

. .= =× ×

= >t fF

172 9 7 3 275840024

0 6428 0 5 OK

χF

F

.

.

. . = = = <

0 5 0 50 6428

0 778 1d

OK

Leff = χFl y = 0.778 × 172.9 = 134.5 mm

F

f L tVRd

yw eff w

M1EdkN kN

. .

. ( . )= =

× ×× = > =−

γ275 134 5 7 3

1 0010 270 79 23 OK

DEFLECTIONThe maximum bending moment due to working load is:

M

g q Lmax

( )

( ) =

+=

+=k k kNm

2 2

8

8 6 8

8112

Elastic resistance:

( )

. maxM

W fMc,Rd el

el,y y

M0

Nmm= =× ×

= × >γ

796 10 275

1 05199 10

36

Hence deflection can be calculated elastically.Deflection due to variable loading (qk) = 6 kNm−1 = 6 Nmm−1

δ

ω

( )

= × = ×

× ×× × ×

=5

384

5

384

6 8 10

210 10 14200 1011

4 3 4

3 4

L

EImm

Assuming the beam supports brittle finishes the maximum permissible deflection is

spanmm

360

8 10

36022

3

=×

= > δ OK

DESIGN BENDING MOMENTDesign action (FEd) = (FEd)udl + (FEd)pl

= (γGgk + γQqk) × span + γQQk

= (1.35 × 6 + 1.5 × 6)6 + 1.5 × 25

= 102.6 + 37.5 = 140.1 kN

Example 9.2 Design of a laterally restrained beam (EC 3)Select and check a suitable beam section using S 235 steel to support the loads shown below. Assume beam is fullylaterally restrained and that it sits on 125 mm bearings at each end.

9780415467193_C09 9/3/09, 4:22 PM387


388


Design bending moment M

F l F lEd

Ed udl Ed pl kNm ( )

( )

.

.

. = + =×

+×

=8 4

102 6 6

8

37 5 6

4133 2

SECTION SELECTION(Refer to Resistance of cross-sections: bending moments)

Assume suitable section belongs to class 1Hence, the minimum required plastic moment of resistance about the major axis (y–y), Wpl,y, is given by

W

Mfpl,y

pl,Rd M0

y

≥γ

Putting Mpl,Rd = MEd and assuming fy = 235 N mm2 (Table 9.4) gives

W

M

fpl,y

pl,Rd M0

y

mm cm . .

. . ≥ =× ×

= × =γ 133 2 10 1 00

235566 8 10 566 8

63 3 3

From the steel table (Appendix B), try 356 × 171 × 45 kg m−1 UB (Wpl,y = 774 cm3)

CHECK STRENGTH CLASSIFICATIONFlange thickness (tf) = 9.7 mmHence, from Table 9.4 fy = 235 N mm−2 as assumed

CHECK SECTION CLASSIFICATION

ε = (235/fy)0.5 = (235/235)0.5 = 1.0

ct f

..

=71 859 7

= 7.41 < 9ε = 9 × 1.0 = 9

(where c = (b − tw − 2r )/2 = (171.0 − 6.9 − 2 × 10.2)/2 = 143.7/2 = 71.85 mm)

ct*

w

.

.=

312 36 9

= 45.3 < 72ε = 72 × 1.0 = 72

Hence from Table 9.5, section belongs to class 1 as assumed.


BendingPlastic moment of resistance, Mpl,Rd, of 356 × 171 × 45 UB is given by

M

W fpl,Rd

pl,y y

M0

.= =

× ×γ

774 10 2351 00

3

= 181.9 × 106 Nmm = 181.9 kNm > MEd (= 133.2 kNm) OK

ShearDesign shear force, VEd, is

V

FEd

Ed kN= = = .

2

140 1

270

9780415467193_C09 9/3/09, 4:22 PM388

389

Design of beams

For class 1 section, design plastic shear resistance, Vpl,Rd, is given by

Vpl,Rd = A v ( / )fy 3 /γM0

where A v = A − 2btf + (tw + 2r )tf > ηhwtw = 1.0 × 332.6 × 6.9 = 2295 mm2

= 57 × 102 − 2 × 171.0 × 9.7 + (6.9 + 2 × 10.2)9.7

= 2647.4 mm2 OK

(hw = h − 2t f = 352.0 − 2 × 9.7 = 332.6 mm)

Hence, Vpl,Rd = 2647.4 ( / )235 3 /1.00

= 359192 N = 359 kN > VEd (= 70 kN) OK

Bending and shearVEd = 0 at mid-span section, hence the moment of resistance of the section is unaffected.

SHEAR BUCKLING RESISTANCE

As

ht

w

w

.

. .

.

. = = < = × =

332 66 9

48 2 72 721 01 0

72εη

, no check on shear buckling is required.

FLANGE INDUCED BUCKLINGAw, area of web = (h − 2tf)tw = (352 − 2 × 9.7) × 6.9 = 2294.9 mm2

Afc, area of compression flange = btf = 171 × 9.7 = 1658.7 mm2

ht

kEf

AA

w

w yf

w

fc

.

. . .

..

.= = ≤ = ××

=332 66 9

48 2 0 3210 10

2352294 91658 7

315 33

OK


WEB BUCKLINGAt an unstiffened end support

k

s chF

s

w

= ++

≤2 6 6

= +

+

=

. .2 6

125 0332 6

4 25

F k E

t

hcr F

w

w

N . . . .

. = = × × × × =0 9 0 9 4 25 210 10

6 9

332 6793370

33

3

m

f bf t1

yf f

yw w

. .

.= =××

=235 171 0235 6 9

24 8

Assuming d F > 0.5

m

ht2

w

f

. . .

. .=

= ×

=0 02 0 02332 69 7

23 52 2


9780415467193_C09 9/3/09, 4:23 PM389


390


Be

F w

yw ws mm = ≤ + = + =

k Etf h

s c2

2125 0 125

=

× × ×× ×

= . 4.25 210 10 6.9

2 235 332.6mm

3 2

271 8

Hence le = 125 mm

B B

By e f

1 e

f

= + +

+tm

tm

2

2

2

= + +

+ = ..

.

. 125 9 724 8

21259 7

23 52

262.8 mm

B By e f 1 = + +t m m2

By = 125 + 9 7 24 8 23 5. . .+ = 192.4 mm

Hence, ly = 192.4 mm

d

BF

y w yw

cr

. .

. .= =× ×

= >t fF

192 4 6 9 235793370

0 627 0 5 OK

χF

F

.

.

. . = = = <

0 5 0 50 627

0 8 1d

OK

Leff = χFly = 0.8 × 192.4 = 153.9 mm

F

f L tVRd

yw eff w

M1EdkN kN

. .

. . ( )= =

× ×× = > =−

γ235 153 9 6 9

1 0510 237 6 703 OK

DEFLECTIONDeflection due to variable uniformly distributed loading qk = 6 kN m−1 = 6 Nmm−1 and variable point load, Q k =25000 N is

Hence, δ2

4 35384 48

= +q L

EIQ L

EIk k

=

× × ×× × × ×

+× × ×

× × × ×

( )

( )

5 6 6 10

384 210 10 12100 10

25 10 6 10

48 210 10 12100 10

3 4

3 4

3 3 3

3 4

= 4 mm + 4.4 mm

= < = =

×=

8

6 10

36016

3

mm allowablespan

360mm OK

9780415467193_C09 9/3/09, 4:23 PM390

391

Design of beams

Example 9.3 Design of a cantilever beam (EC 3)A cantilever beam is needed to resist the loading shown below. Select a suitable UB section in S 275 steel to satisfybending and shear criteria only assuming full lateral restraint.

DESIGN BENDING MOMENT AND SHEAR FORCEDesign action (FEd) = γGgk + γQqk

= 1.35 × 500 + 1.5 × 350 = 1200 kN

Design bending moment at A, MEd, is

M

F lEd

Ed kNm .

= = × =2

12001 5

2900

Design shear force at A, VEd, is

VEd = FEd = 1200 kN

SECTION SELECTIONSince the cantilever beam is relatively short and subject to fairly high shear forces, the bending capacity or the shearstrength of the section may be critical and both factors will need to be considered in order to select an appropriatesection.

Hence, plastic moment of section (assuming beam belongs to class 1), Wpl, must exceed the following to satisfybending:

W

M

fpl,y

pl,Rd M0

y

mm cm .

≥ =× ×

= × =γ 900 10 1 00

2653396 10 3396

63 2 3

The shear area of the section, A v, must exceed the following to satisfy shear:

A

V

fν

γ

( / )

.

( / ) ≥ =

× ×=pl,Rd M0

y

mm3

1200 10 1 00

265 37843

32

where A v = A − 2btf + (tw + 2r)tf > ηhwtw

Hence, try 610 × 305 × 149 kg m−1 UB section

CHECK STRENGTH CLASSIFICATIONFlange thickness (tf) = 19.7 mmHence from Table 9.4, fy = 265 N mm−2 as assumed

CHECK SECTION CLASSIFICATIONε = (235/fy)

0.5 = (235/265)0.5 = 0.94

ct f

..

=129 9519 7

= 6.59 < 9ε = 9 × 0.94 = 8.46

9780415467193_C09 9/3/09, 4:23 PM391


392

Example 9.3 continued(where c = (b − tw − 2r )/2 = (304.8 − 11.9 − 2 × 16.5)/2 = 259.9/2 = 129.95 mm)

ct*

w

.

.=

537 211 9

= 45.14 < 72ε = 72 × 0.94 = 67.68



BendingPlastic moment of resistance of 610 × 305 × 149 UB about the major axis (y-y) is given by

M

W fpl,Rd

pl,y y

M0

.= =

× ×γ

4570 10 2651 00

3

= 1211 × 106 Nmm = 1211 kNm > MEd (= 900 kNm) OK

ShearFor Class 1 section, design plastic shear resistance, is given by

Vpl,Rd = A v(fy/ 3 )/γM0

where A v = A − 2btf + (tw + 2r)t f > ηhwtw = 1.0 × 570.2 × 11.9 = 6785.38 mm2

= 190 × 102 − 2 × 304.8 × 19.7 + (11.9 + 2 × 16.5)19.7

= 7875.4 mm2 OK

(hw = h − 2tf = 609.6 − 2 × 19.7 = 570.2 mm)

Hence, Vpl,Rd = 7875.4 ( / )265 3 /1.00 = 1204.9 × 103 N

= 1204.9 kN > VEd (= 1200 kN) OK

Bending and shearSince

VEd = 1200 kN > 0.5Vpl,Rd = 0.5 × 1204.9 = 602.5 kN

the section is subject to a ‘high shear load’ and the design resistant moment of the section (about the major axis)should be reduced to My,V,Rd which is given by

M

W A t fy,V,Rd

pl,y w2

w y

M0

( / )

( . . / . )

.=

−=

× − × ×ργ

4 4570 10 0 984 6785 38 4 11 9 2651 00

3 2

= 958.8 × 106 Nmm = 958.8 kNm > MEd (= 900 kNm) OK

where

ρ = (2VEd/Vpl,Rd − 1)2 = (2 × 1200/1204.9 − 1)2 = 0.984

Aw = hwtw = 570.2 × 11.9 = 6785.38 mm2

Selected UB section, 610 × 305 × 149, is satisfactory in bending and shear.

9780415467193_C09 9/3/09, 4:24 PM392

393

Design of beams

Example 9.4 Design of a beam with stiffeners (EC 3)The figure shows a simply supported beam and cantilever with uniformly distributed loads applied to it. Using gradeS 275 steel and assuming full lateral restraint, select and check a suitable beam section.

DESIGN SUMMARYPermanent action = 200 kN m−1

Variable action = 100 kN m−1

Grade S 275 steelFull lateral restraint

ULTIMATE LIMIT STATE3 load cases:(a) 1.35Gk + 1.5Qk on span 1

1.35Gk + 1.5Qk on span 2(b) 1.35Gk + 1.5Qk on span 1

1.0Gk + 0.0Qk on span 2(c) 1.0Gk + 0.0Qk on span 1

1.35Gk + 1.5Qk on span 2

Static equilibrium (refer to Table 8.6)Overturning moment = (1.1 × 200 + 1.5 × 100)2.5 × 1.25 = 1156.25 kNm (load case (c))Restorative moment = (0.9 × 200)5 × 2.5 = 2250 kNm > 1156.25 kNm OK

SHEAR AND BENDINGShear force and bending moment diagrams for the 3 load cases are shown below.

9780415467193_C09 9/3/09, 4:24 PM393


394

Example 9.4 continuedSECTION SELECTIONFor class 1 section

W

M

fpl,y

pl,Rd M0

y

mm cm .

. ≥ =× ×

= × =γ 1313 10 1 00

2654 955 10 4955

66 2 3

Try 762 × 267 × 173 UB section

STRENGTH CLASSIFICATIONFlange thickness = 21.6 mm, steel grade S 275. Hence from Table 9.4, fy = 265 N mm−2.


ε = (235/fy)1/2 = (235/265)1/2 = 0.94

ct f

..

=109 721 6

= 5.08 < 9ε = 9 × 0.94 = 8.46

(where c = (b − tw − 2r )/2 = (266.7 − 14.3 − 2 × 16.5)/2 = 219.4/2 = 109.7 mm)

ct∗

=w

.

.685 814 3

= 47.9 < 72ε = 72 × 0.94 = 67.7


RESISTANCE OF CROSS-SECTIONS

Shear resistanceFor class 1 section, design plastic shear resistance, is given by

Vpl,Rd = A v ( / )fy 3 /γMO

where A v = A − 2btf + (tw + 2r)tf > ηhwtw = 1.0 × 718.8 × 14.3 = 10278.8 mm2

= 220 × 102 − 2 × 266.7 × 21.6 + (14.3 + 2 × 16.5)21.6

= 11500.2 mm2 OK

(hw = h − 2tf = 762 − 2 × 21.6 = 718.8 mm)

Hence, Vpl,Rd = 11500.2 ( / )265 3 /1.00 = 1759.5 × 103 N

= 1759.5 kN > VEd (= 1313 kN) OK

Bending and shear

VEd = 1313 kN = 0.746Vpl,Rd > 0.5Vpl,Rd

Hence, beam is subject to HIGH SHEAR LOAD

M

W A t fy,V,Rd

pl,y w2

w y

M0

mm

( / )

( . . / . )

.=

−=

× − × ×ργ

4 6200 10 0 243 10278 8 4 14 3265

1 00

3

= 1524 × 106 Nmm = 1524 kNm > MEd (= 1313 kNm) OK

9780415467193_C09 9/3/09, 4:25 PM394

395

Design of beams

where

ρ = (2VEd/Vpl,Rd − 1)2 = (2 × 1313/1759.5 − 1)2 = 0.243

Aw = hwtw = 718.8 × 14.3 = 10278.8 mm2

Shear buckling resistance

As will be seen later, the web needs stiffeners. Hence, provided c ∗/tw for the web is less than

31η

ε τk (clause 5.1,

EC 3–5) no check is required for resistance to shear buckling.

kτ = 5.34 (assuming stiffeners needed only at supports)ε = 0.94η = 1.2

Hence,

31 311 2

0 94 5 34 56 1 48 0η

ε τk c t .

. . . / ( . )= × × = > ∗ =w OK

Flange-induced buckling

ht

kEf

AA

w

w yf

w

fc

.

. . .

.= = ≤ = ×

×=

718 814 3

50 3 0 3210 10

265102795761

317 63

whereAw, area of web = (h − 2tf)tw = (762 − 2 × 21.6) × 14.3 = 10279 mm2

A fc, area of compression flange = btf = 266.7 × 21.6 = 5761 mm2


Resistance of web to transverse forcesThe web capacity at supports A and C needs checking but by inspection it would seem that the web resistance at Cis almost certainly more critical. Therefore only the calculations for this support are presented.


Stiff bearing length, ss ≈ 2t f + tw + r = 2 × 36.6 + 21.5 + 24.1 = 118.8 mmSince the load is applied through the flange and is resisted by shear forces in the web (i.e. load type (a) in Fig. 9.3),kF is given by

k

haFw = +

6 22

9780415467193_C09 9/3/09, 4:25 PM395


396

Example 9.4 continuedAs there are no web stiffeners, a = ∞ and therefore hw/a = 0 and kF = 6.

F k E

t

hcr F

w

w

N . . .

. . = = × × × × =0 9 0 9 6 210 10

14 3

718 84613314 8

33

3

m

f bf t1

yf f

yw w

. .

= =××

=265 266 7275 14 3

18

Assuming d F > 0.5

m

ht2

w

f

. . ..

.=

= ×

=0 02 0 02718 821 6

22 12 2

By s f 1 = + + +( )s t m m2 1 2

= + × + +( ) . . .118 8 2 36 6 1 18 22 1

= 655.5 mm

d

BF

y w yw

cr

. .

. . .= =

× ×= >

t f

F

655 5 14 3 275

4613314 80 75 0 5 OK

χF

F

.

.

. . = = = <

0 5 0 5

0 750 67 1

dOK

Leff = χFl y = 0.67 × 655.5 = 439 mm

F

f L tRd

yw eff w

M1

.

. = =

× ×× −

γ275 439 14 3

1 0010 3

= 1726 kN < FEd (= 2363 kN at C, load case (a)) Not OK

Where the member is subjected to a concentrated transverse force acting on the compression flange and bending, e.g.support C, clause 7.2 of EC 3–5 recommends the following additional check:

η2 + 0.8η1 ≤ 1.4

where

η

γ1

6

6

1313 106 2 10 265 1 00

0 80 ( / )

. ( / . ) .= =

××

=M

W fEd

pl,y y M1

η

γ2

32363 10

439 14 3 275 1 001 37

( / )

. ( / . ) .= =

××

=F

L t fEd

eff w yw M1

η2 + 0.8η1 = 1.37 + 0.8 × 0.80 = 2.0 > 1.4

Hence, this is also not satisfied. Stiffeners are therefore required at A and C with the stiffener at C being the morecritical.

9780415467193_C09 9/3/09, 4:26 PM396

397

Design of beams

Example 9.4 continuedStiffener design at C (Refer to Cl. 9, EC 3–5)Assume a stiffener thickness, S t, of 15 mm and width, Sw, of 250mm (see figure).

15

When checking the buckling resistance, the effective cross-section of a stiffener should be taken as including a widthof web plate equal to 30εtw, arranged with 15εtw = 15 × (235/275)0.514.3 = 198 mm each side of the stiffener.Buckling length, l ≥ 0.75hw = 0.75 × 718.8 = 539 mmRadius of gyration of stiffened section, ix, is

ix = (I/A)1/2 = [(15 × 2503/12)/(2 × 198 × 14.3 + 15 × 250)]1/2

= (19531250/9412.8)1/2 = 45.6 mm

λ = l/i = 539/45.6 = 11.8

λ1 = 93.9ε = 93.9 × ( / )235 fy = 93.9 × ( / )235 275 = 86.8

d = λ/λ1

= 11.8/86.8 = 0.136 < 0.2

From Fig. 6.4 of EC 3–1, reduction factor χc = 1.0Design buckling resistance, Nb,Rd, is given by (6.3.1.1(3))

N

AfNb,Rd

y

M1

kN kN . .

. . ( )= =

× ×= × = >

χγ

1 0 9412 8 2751 00

2 558 10 2558 23636 OK

Hence double-sided stiffeners made of two plates 118 × 15 is suitable at support C. A similar stiffener should beprovided at support A.

SERVICEABILITY LIMIT STATEFor a simply supported span

δ

ω = ×

5384

4LEI

where ω is the unfactored imposed load = 100 kN m−1 = 100 N mm−1

δ

ω

( ) .

= × = ×× ×

× × ×= = <

5384

5384

100 5 10210 10 2 05 10

22580 360

4 3 4

3 9

LEI

mmL L

OK

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398

9.11.2 LATERAL TORSIONAL BUCKLING OFBEAMS (CL. 6.3.2, EC 3)

In order to prevent the possibility of a beam failuredue to lateral torsional buckling, the designer needsto ensure that the buckling resistance, Mb,Rd exceedsthe design moment, MEd, i.e.

MM

Ed

b,Rd

.≤ 1 0 (9.36)

The buckling resistance is determined using equa-tion 9.45 and is a function of

(1) the elastic critical moment, Mcr

(2) the buckling factor, χLT

The following discusses how the design values ofthese parameters are assessed.

9.11.2.1 Elastic critical moment, McrEC 3 provides no detailed information on how tocalculate the design elastic critical moment of beamsbut rather leaves it to the designer to use some-thing appropriate. No reason for this omission isprovided in the code and on that basis the follow-ing formula for the elastic critical moment, Mcr,taken from the prestandard (i.e. ENV 1993–1–1,1992), is recommended.

M

EIL

II

L GIEIcr

z

cr

w

z

cr t

z

.

= +

ππ

2

2

2

2

0 5

(9.37)

whereLcr = length of beam between points which have

lateral restraintIz = second moment of area about the minor

axis (z–z)Iy = second moment of area about the major

axis (y–y)It = torsional constantIw = warping constantE = modulus of elasticity (210000 N mm−2)

G = shear modulus =

E2 1

2100002 1 0 3( )

( . )+

=+ν

= 80769 Nmm−2 ≈ 81000 Nmm−2

Equation 9.37 is based on the following assumptions:

1. The beam is of uniform section with equal flanges.2. Beam ends are simply supported in the lateral

plane and prevented from lateral movement andtwisting about the longitudinal axis but are freeto rotate on plan.

3. The section is subjected to equal and oppositein plane end moments.

4. The loads are not destabilising (section 4.8.11.1).

9.11.2.2 Buckling factor χLTχLT is the reduction factor for lateral torsional buck-ling. Two methods of calculating χLT are providedin EC 3 as follows:

(1) The so called ‘general case’ mentioned inclause 6.3.2.2 is applicable to all members ofconstant cross-section.

(2) The approach detailed in clause 6.3.2.3 isapplicable to rolled sections only.

(a) General case (Cl. 6.3.2.2, EC 3). Accordingto the general case χLT is given by

χLT

LT LT2

LT2

=+ −

1

Φ Φ d ≤ 1 (9.38)

where

ΦLT LT LT LT2 . [ ( . ) ]= + − +0 51 0 2α d d (9.39)

in whichαLT is an imperfection factor (Table 9.7)

dLT

y y

cr

=W fM

(9.40)

where

Wy = Wpl,y for class 1 or 2 sections= Wel,y for class 3 sections= Weff,y for class 4 sections

Recommended values for imperfection factorsfor lateral torsional buckling, αLT, are given inTable 9.7. Whichever of the buckling curves is useddepends on the type and size of section used asindicated in Table 9.8.

Table 9.8 Recommended values for lateraltorsional buckling curves for rolled sections(based on Table 6.4, EC 3)

Cross-section Limits Buckling curve

Rolled I-sections h/b ≤ 2 ah/b > 2 b

Table 9.7 Imperfection factors (Table 6.3,EC 3)

Buckling curve a b c d

Imperfection factor αLT 0.21 0.34 0.49 0.76

9780415467193_C09 9/3/09, 4:26 PM398

399

Design of beams

(b) Buckling factor χχχχχLT for rolled sections(Cl. 6.3.2.3, EC 3). An alternative method ofcalculating the reduction factor for lateral torsionalbuckling χLT for rolled sections only is described inclause 6.3.2.3. Here χLT is given by

χ

βLTLT LT

2LT2

=+ −

1

Φ Φ d

but χLT ≤ 1 and χLT

LT2

≤1

d(9.41)

where

ΦLT LT LT LT,0 LT2 . [ ( ) ]= + − +0 51 α βd d d (9.42)

in whichαLT is an imperfection factorDLT,0 = 0.4 (for rolled sections, clause 2.17 of UK

National Annex to EC 3)β = 0.75 (for rolled sections, clause 2.17 of

UK National Annex to EC 3)

The value of the imperfection factor used inequation 9.42 is again taken from Table 9.7. Butthe appropriate buckling curve for the cross-section is determined from Table 9.9.

In order to take account of the bending momentcurve between points of lateral restraint the reduc-tion factor χLT may be modified as follows:

χ

χLT,mod

LT .= ≤f

1 0 (9.43)

where

f = 1 − 0.5(1 − kc)[1 − 2.0(dLT − 0.8)2] ≤ 1.0(9.44)

in which

k

Cc =

1

1

(9.45)

where

C1 =

M

Mcr

cr

for the actual bending moment diagram

for a uniform bending moment diagram

obtained from Table 9.10.

9.11.2.3 Buckling resistance Mb,Rd (Cl. 6.3.2.1, EC 3)The design buckling resistance moment of a later-ally unrestrained beam is given by

M W

fb,Rd LT y

y

M

= χγ 1

(9.45)

whereχLT is the reduction factor for lateral torsional

bucklingWy is the section modulus as follows

= Wpl,y for class 1 or 2 sections= Wel,y for class 3 sections= Weff,y for class 4 sections

fy is the yield strength (Table 9.4)γM1 is the partial safety factor for buckling = 1.0

(section 9.9.3)

Table 9.9 Recommended values for lateraltorsional buckling curves for rolled sections(based on clause 2.17 of the UK National Annexto EC 3)

Cross-section Limits Buckling curve

Rolled I-sections h/b ≤ 2 b2 ≤ h/b < 3.1 ch/b > 2 d

9780415467193_C09 9/3/09, 4:27 PM399


400

Table 9.10 Values of C1 for various end moment or transverse loadingcombinations (based on Tables F.1.1 and F.1.2, ENV EC 3)

Loading and support conditions Bending moment diagram ψ C1

+1 1.000

0 1.879

−1 2.752

– 1.132

– 1.285

– 1.365

– 1.565

– 1.046

9780415467193_C09 9/3/09, 4:27 PM400

401

Example 9.5 Analysis of a beam restrained at supports (EC 3)Assuming that the beam in Example 9.1 is only laterally and torsionally restrained at the supports, determine whethera 356 × 171 × 51 kg m−1 UB section in S275 steel is still suitable.

SECTION PROPERTIESFrom steel tables (Appendix B)

Depth of section, h = 355.6 mmWidth of section, b = 171.5 mmThickness of flange, tf = 11.5 mmSecond moment of area about the minor axis, Iz = 968 × 104 mm4

Radius of gyration about z–z axis, iz = 38.7 mmElastic modulus about the major axis, Wel,y = 796 × 103 mm3

Plastic modulus about the major axis Wpl,y = 895 × 103 mm3

Warping constant, Iw = 286 × 109 mm6

Torsional constant, It = 236 × 103 mm4

Yield strength of S275 steel, fy = 275 N mm−2

Shear modulus, G = 81000 N mm−2

Method for uniform members in bending – General case (Cl. 6.3.2.2)

Elastic critical moment, McrLength of beam between points which are laterally restrained, Lcr = 8000 mm

M

EIL

II

L GIEIcr

z

cr

w

z

cr t

z

.

= +

ππ

2

2

2

2

0 5

=

× × × ××

××

+× × ×

× × × ×

= ×

( )

.

.π

π

2 3 4

3 2

9

4

2 3

2 3 4

0 5

7210 10 968 108 10

286 10968 10

8000 81000 236 10210 10 968 10

9 4318 10 Nmm

For class 1 sections Wy = Wpl,y = 895 × 103 mm3

dLT

y y

cr

. .= =

× ××

=W fM

895 10 2759 4318 10

1 6153

7

ΦLT = 0.5[1 + αLT(dLT − 0.2) + d2LT] = 0.5[1 + 0.34(1.615 − 0.2) + 1.6152] = 2.045

χLT

LT LT2

LT2

. . .

. .=+ −

=+ −

= <1 1

2 045 2 405 1 6150 261 1 0

2 2Φ Φ dOK

Mb,Rd = χLTWyfy/γM1

= 0.261 × 895 × 103 × 275/1.00 = 64.2 × 106 Nmm

= 64.2 kNm < MEd = 158.4 kNm (Example 9.1)

Since buckling resistance of the beam (= 64.2 kNm) is less than the design moment (MEd = 158.4 kNm), the beamsection is unsuitable

MEd = 158.4 kN m

Design of beams

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402

Example 9.6 Analysis of a beam restrained at mid-span and supports(EC 3)

Repeat Example 9.5, but this time assume that the beam is laterally and torsionally restrained at mid-span and at thesupports.

SECTION PROPERTIESSee Example 9.5

ELASTIC CRITICAL MOMENT, Mcr

Length of beam between points which are laterally restrained, L cr = 4000 mm.

M

EIL

II

L GIEIcr

z

cr

w

z

cr t

z

.

= +

ππ

2

2

2

2

0 5

=

× × × × ××

+× × ×

× × × ×

= ×

. .

ππ

2 3 4

2

9

4

2 3

2 3 4

0 5

8210 10 968 104000

286 10968 10

4000 81000 236 10210 10 968 10

2 6538 10 Nmm


dLT

y y

cr

. .= =

× ××

=W fM

895 10 2752 6538 10

0 9633

8

LATERAL BUCKLING RESISTANCE

Using the provision of Cl. 6.3.2.2 (i.e. General case)

hb

..

.= =355 6171 5

2 07. From Table 9.8 use buckling curve b ⇒ α LT = 0.34 (Table 9.7)

ΦLT = 0.5[1 + αLT(d LT − 0.2) + d2LT] = 0.5[1 + 0.34(0.963 − 0.2) + 0.9632] = 1.093

χLT

LT LT2

LT2

. . .

. .=+ −

=+ −

= <1 1

1 093 1 093 0 9630 643 1 0

2 2Φ Φ dOK

Mb,Rd = χLTWyfy/γM1

= 0.643 × 895 × 103 × 275/1.00 = 158.3 × 106 Nmm

= 158.3 kNm ≈ MEd = 158.4 kNm (Example 9.1)

Using the provision of clause 6.3.2.3 (i.e. procedure for rolled I or H sections)From above h/b = 2.07 > 2. From Table 9.9 use buckling curve c ⇒ αLT = 0.49 (Table 9.7)

ΦLT = 0.5[1 + αLT (d LT − d LT,0) + βd2LT] = 0.5[1 + 0.49(0.963 − 0.4) + 0.75 × 0.9632] = 0.986

9780415467193_C09 9/3/09, 4:28 PM402

403

χ

βLTLT LT

2LT2

=+ −

1

Φ Φ d

=

+ − ×= < < = =

. . . . .

. .

10 986 0 986 0 75 0963

0 661 1 01 1

0 9631 04

2 2 dLT2

OK

Ratio of end moments for spans A & B is ψ, where ψ = 0/158.4 = 0. Hence from Table 9.10(b) C1 = 1.879.

k

Cc .

.= = =1 1

1 8790 729

1

f = 1 − 0.5(1 − kc)[1 − 2.0(dLT − 0.8)2] = 1 − 0.5(1 − 0.729)[1 − 2 × (0.963 − 0.8)2] = 0.872

χ

χLT,mod

LT ..

. .= = = <f

0 6610 872

0 758 1 0 OK

Mb,Rd = χLT,modWyfy/γM1

= 0.758 × 895 × 103 × 275/1.00 = 186.6 × 106 Nmm

= 186.6 kNm > MEd = 158.4 kNm (Example 9.1)

This approach gives a higher estimate of the buckling resistance of rolled sections which will generally be the case.


9.12 Design of columnsThe design of columns is largely covered withinchapter 6 of EC 3. However, unlike BS 5950, EC3 does not include the design of cased columns,which is left to Eurocode 4 on Steel ConcreteComposite Structures.

The following sub-sections will consider EC 3requirements in respect of the design of

(i) compression members;(ii) members resisting combined axial load and

bending;(iii) columns in simple construction;(iv) simple column baseplates.

9.12.1 COMPRESSION MEMBERSCompression members (i.e. struts) should bechecked for

(1) resistance to compression(2) resistance to buckling

9.12.1.1 Compression resistance ofcross-sections (Cl. 6.2.4., EC 3)

For members in axial compression, the design valueof the compression force NEd at each cross sectionshould satisfy

NN

Ed

c,Rd

.≤ 1 0 (9.46)

where Nc,Rd is the design compression resistance ofthe cross-section, taken as

(a) the design plastic resistance of the gross cross-section

N

Afc,Rd

y

M

=γ 0

(9.47)

(for class 1, 2, and 3 cross-sections);(b) the design local buckling resistance of the

effective cross-section

N

A fc,Rd

eff y

M

=γ 0

(9.48)

(for class 4 cross-sections)

where Aeff is the effective area of section.

9.12.1.2 Buckling resistance of members(Cl. 6.3, EC 3)

A compression member should be verified againstbuckling as follows:

NN

Ed

b,Rd

.≤ 1 0 (9.49)

whereNEd is the design value of the compression forceNb,Rd is the design buckling resistance of the

compression member

Design of columns

9780415467193_C09 9/3/09, 4:28 PM403


404

The design buckling resistance of a compressionmember should be taken as:

N

Afb,Rd

y

M

=χγ 1

for class 1, 2 and 3 cross-sections (9.50)

The equivalent equation in BS 5950 is Pc = pc A

N

A fb,Rd

eff y

M

=χ

γ 1

for class 4 cross-sections (9.51)

whereχ is the reduction factor for the relevant buckling

mode, given by

χ

.=

+ −≤

11 0

2 2Φ Φ d(9.52)

in which

Φ = 0.5[1 + α(d − 0.2) + d2] (9.53)

whereα is an imperfection factor from Table 9.11.

Table 9.12 indicates which of the bucklingcurves is to be used.

d is the non-dimensional slenderness, generallygiven by

d =

λλ1

For class 1, 2 and 3 cross-sections

d = = ×

AfN

Li

y

cr

cr 1

1λ(9.54)

and for class 4 cross-sections

d

/= = ×

A fN

Li

A Aeff y

cr

cr eff

λ1

(9.55)

whereNcr is the elastic critical buckling load for the

relevant buckling mode based on the grosscross-sectional properties

λ π1 = =

Efy

93.9 (9.56)

ε =

235fy

( fy in N mm−2) (9.57)

In equations (9.54) and (9.55) Lcr is the bucklinglength of the member in the plane under con-sideration and i is the radius of gyration about therelevant axis. Values of radius of gyration for UBand UC sections can be obtained from steel tables(Appendix B). However, EC 3 provides little infor-mation on how to determine Lcr, which is in fact

Table 9.12 Selection of buckling curve for a cross-section(based Table 6.2, EC 3)

Cross-section Limits Buckling Bucklingabout axis curve

Rolled sections h /b > 1.2t f ≤ 40 mm y–y a

z–z b

40 mm < t f ≤ 100 mm y–y bz–z c

h/b ≤ 1.2t f ≤ 100 mm y–y b

z–z c

t f > 100 mm y–y dz–z d

Table 9.11 Imperfection factors for bucklingcurves (Table 6.1, EC 3)

Buckling curve ao a b c d

Imperfection factor α 0.13 0.21 0.34 0.49 0.76

9780415467193_C09 9/3/09, 4:28 PM404

405

equivalent to effective length, LE, in BS 5950. Theguidance provided in the prestandard yielded val-ues of buckling length which were identical to BS5950 effective lengths, and therefore it would seemreasonable to use either the guidance in AnnexE of the prestandard or Table 22 of BS 5950 (Table4.15) to determine Lcr.

9.12.2 MEMBERS SUBJECT TO COMBINEDAXIAL FORCE AND BENDING

The design approach for members subject tocombined axial (compression) force and bendingrecommended in EC 3 is based on the equivalentmoment factor method, which necessitates that thefollowing ultimate limit states are checked:

(1) resistance of cross-sections to the combinedeffects

(2) buckling resistance of member to the combinedeffects.

9.12.2.1 Resistance of cross-sections –bending and axial force (Cl. 6.2.9, EC 3)

Classes 1 and 2 cross-sections. For class 1 and2 cross-sections subject to an axial force anduniaxial bending the criterion to be satisfied inthe absence of shear force is

MM

Ed

N,Rd

≤ 1 (9.58)

where MN,Rd is the reduced design plastic resistancemoment allowing for the axial force, NEd. Forbending about the y–y axis no reduction is neces-sary provided that the axial force does not exceedhalf the plastic tension resistance of the web (i.e.

0 5

0

. w w y

M

h t fγ

), or a quarter of the plastic tension

resistance of the whole cross-section (i.e. 0.25Npl,Rd),whichever is the smaller.

For larger axial loads the following approxi-mations can be used for standard rolled I and Hsections

M M

na

MN,y,Rd pl,y,Rd pl,y,Rd ( )

( . ) =

−−

≤1

1 0 5(9.59)

For n ≤ a: MN,z,Rd = Mpl,z,Rd (9.60)

For n > a:

M Mn a

aN,z,Rd pl,z,Rd

= −−−

11

2

(9.61)

wheren = NEd/Npl,Rd

a = (A − 2btf)/A ≤ 0.5

For bi-axial bending, the following approxi-mate criterion can be used:

MM

MM

y,Ed

N,y,Rd

z,Ed

N,z,Rd

+

≤

α β

1 (9.62)

where for I and H sections α = 2 and β = 5n butβ ≥ 1.

As a further conservative approximation for class1, 2 and 3 cross-sections the following may beused (clause 6.2.1(7)):

NN

MM

MM

Ed

Rd

y,Ed

y,Rd

z,Ed

z,Rd

+ + ≤ 1 (9.63)

whereNRd = A fy/γM0

My,Rd = Wy fy /γM0

Mz,Rd = Wz fy /γM0

Class 3 cross-sections. In the absence of a shearforce, class 3 cross-sections will be satisfactory ifthe maximum longitudinal stress does not exceedthe design yield strength, i.e.

σ

γx,Edy

M

≤f

0

For cross-sections without fastener holes, thisbecomes (clauses 6.2.4 and 6.2.5)

NAf

MW f

MW f

Ed

y M

y,Ed

el,y y M

z,Ed

el,z y M/

/

/

γ γ γ0 0 0

1+ + ≤ (9.64)

Class 4 cross-sections. For class 4 cross-sectionsthe above approach should also be used, but calcu-lated using effective, rather than actual, widths ofcompression elements (Cl. 6.2.9.3(2)).

9.12.2.2 Buckling resistance of members –combined bending and axialcompression (Cl. 6.3.3, EC 3)

Again this is presented in a rather cumbersomemanner in EC 3. We will confine our remarks toclass 1 and 2 members. Members which are sub-jected to combined bending and axial load shouldsatisfy the following

NN

kMM

kM

MEd

y Rk Myy

y,Ed

LT y,Rk Myz

z,Ed

z,Rk Mχ γ χ γ γ/

/

/

1 1 1

1+ + ≤

(9.65)

NN

kMM

kM

MEd

z Rk Mzy

y,Ed

LT y,Rk Mzz

z,Ed


/

/

1 1 1

1+ + ≤

(9.66)

Design of columns

9780415467193_C09 11/3/09, 11:18 AM405


406

whereNEd, MyEd are respectively the design valuesand Mz,Ed of the compression force and the

maximum moments about the y–yand z–z axis along the member

χy and χz are the reduction factors due toflexural buckling as discussed in9.12.1.2

χLT is the reduction factor due to lateraltorsional buckling as discussed in9.11.2.2

kyy, kyz, kzy are the interaction factors (seeand kzz below)NRk = Afy (for class 1 and 2 sections)My,Rk, Mz,Rk are respectively, Wpl,y fy, Wpl,z fy

(for class 1 and 2 sections)

Interaction factors. Two alternative methodsfor calculating values of the interaction factorsused in equations 9.65 and 9.66 are presented inEC 3. Both methods are based on second-orderin-plane elastic stability theory and have beenvalidated using the same set of results of numericalsimulations from finite element analysis and lab-oratory data. However, the emphasis in Method 1appears to be on transparency of structural beha-viour and precision whereas in the case of Method 2some allowance for practical convenience has alsobeen made. Consequently, only the formulationfor Method 2, contained within Annex B, is pre-sented here as it is somewhat simpler and therefore

Table 9.13 Equivalent uniform moment factors Cm for bracedcolumns with linear applied moments between restraints (Table B3,EC 3)

Moment diagram Range Cmy, Cmz and CmLT

−1 ≤ ψ ≤ 1 0.6 + 0.4ψ ≥ 0.4

Cmy, Cmz and CmLT should be obtained according to the bending momentdiagram between the relevant braced points as follows:

Moment factor bending axis points braced in directionCmy y–y z–zCmz z–z y–yCmLT y–y y–y

more amenable to hand calculations. The expres-sions relevant to the design of classes 1 and 2 rolledsections, susceptible to torsional deformation, areas follows:

k C

NNyy my y

Ed

y Rk M

[ . ]/

= + −

1 0 2

1

dχ γ

≤ +

.

/C

NNmy

Ed

y Rk M

1 0 81χ γ

(9.67)

kyz = 0.6kzz (9.68)

k

CN

Nzyz

mLT

Ed

z Rk M

. .

/

= −−

×10 1

0 25 1

dχ γ

≥ −

−×

. .

/

10 1

0 25 1CN

NmLT

Ed

z Rk Mχ γ(9.69)

k C

NNzz mz z

Ed

z Rk M

[ . ]/

= + −

1 2 0 61

dχ γ

≤ +

./

CN

NmzEd

z Rk M

1 1 41χ γ

(9.70)

whereCmy, Cmz and Cm,LT are the equivalent uniformmoment factors determined from Table 9.13 andaccount for the shape of the bending moment dia-gram between the relevant points of restraint.

M

ψM

9780415467193_C09 9/3/09, 4:29 PM406

407

9.12.4.1 Bearing pressure and strengthThe aim of design is to ensure that the bearingpressure does not exceed the bearing strength ofthe concrete, fj, i.e.

bearing pressure =

NA

fEd

effj ≤ (9.72)

whereNEd is the design axial force on columnAeff is the area in compression under the base

plate.

The bearing strength of concrete foundations canbe determined using

fj = β jk j fcd (9.73)

wherefcd is the design value of the concrete cylinder

compressive strength of the foundationdetermined in conformity with Eurocode2 = αcc fck/γC

βj is the joint (concrete) coefficient which maygenerally be taken as 2/3

kj is the concentration factor which maygenerally be taken as 1.0.

Figure 9.7 shows the effective bearing areas underaxially loaded column base plates. The additionalbearing width x is given by

x t

ff

/

=

y

j M3 0

1 2

γ(9.74)

wheret is the thickness of the base platefy is the yield strength of base plate materialfj is the bearing strength of the foundations.

9.12.4.2 Resistant momentTo prevent bending failure, the bending moment inthe baseplate, msd, must not exceed the resistancemoment, mRd:

msd < mRd (9.75)

The bending moment in the base plate is given by

m sd = (x2/2)NSd/Aeff (9.76)

The resistance moment is given by

m

t fRd

y

M

=2

06γ(9.77)

Design of columns

The complexity of the above expressions largelyarises from the fact that they model beam-column behaviour more closely than the modelspresented in BS 5950 and indeed ENV EC 3with regard to:

1. torsional effects2. buckling effects3. instability effects.

9.12.3 COLUMNS IN SIMPLE CONSTRUCTIONA simplified conservative approach to the design ofcolumns in simple construction has been developedby the Steel Construction Institute (SN 048b)which avoids the calculation of the interactionfactors discussed above. It is suitable for classes1–3, hot rolled UB and UC sections and involveschecking the following expression is satisfied

NN

MM

MM

Ed

b,z,Rd

y,Ed

b,Rd

z,Ed

z,Rd

. + + ≤1 5 1 (9.71)

whereMb,Rd = χLTWy fy/γM1 (from clause 6.3.2.1, EC 3)Mz,Rd = Wpl,z fy/γM1

Nb,z,Rd =

χγz y

M

Af

1

in whichχz is determined using equations (9.52)–(9.57)

and Tables 9.11 and 9.12.

As no guidance is provided in EC 3 to evaluatethe design moments acting on columns in simpleconstruction, i.e. My,Ed and Mz,Ed, it is recommendedthat the rules given in clause 4.7.7 of BS 5950concerning values of eccentricities to the columnface of beam reactions are used (section 4.9.4).

9.12.4 DESIGN OF COLUMN BASE PLATES(L.1, EC 3)

The following is based on the advice given in AnnexL of the ENV version of EC 3.

Generally, column base plates should be checkedto ensure

(1) the bearing pressure does not exceed the designbearing strength of the foundations

(2) the bending moment in the compression ortension region of the base plate does not exceedthe resistance moment.

9780415467193_C09 9/3/09, 4:29 PM407


408

Example 9.7 Analysis of a column resisting an axial load (EC 3)Check the suitability of the 203 × 203 × 60 kg m−1 UC section in S275 steel to resist a design axial compression forceof 1400 kN. Assume the column is pinned at both ends and that its height is 6m.

SECTION PROPERTIESFrom steel tables (Appendix B)Area of section, A = 7580 mm2

Thickness of flange, tf = 14.2 mmRadius of gyration about the major axis (y–y), i y = 89.6mmRadius of gyration about the minor axis (z–z), i z = 51.9mm

NEd = 1400 kN

Fig. 9.7 Area in compression under base plate: (a) general case; (b) short projection; (c) large projetion (Fig. L. 1, ENVEC3).

9780415467193_C09 9/3/09, 4:29 PM408

409

STRENGTH CLASSIFICATIONFlange thickness = 14.2 mm, steel grade S275Hence from Table 9.4, fy = 275 N mm−2


ε = (235/fy)0.5 = (235/275)0.5 = 0.92

ct f

..

=87 7514 2

= 6.18 < 9ε = 9 × 0.92 = 8.28

(where c = (b − tw − 2r)/2 = (205.2 − 9.3 − 2 × 10.2)/2 = 175.5/2 = 87.75 mm)Also

ct

dt

∗= =

w

.

.160 9

9 3 = 17.3 < 33ε = 33 × 0.92 = 30.36

Hence from Table 9.5, section belongs to class 1.

RESISTANCE OF CROSS-SECTION – COMPRESSIONDesign resistance for uniform compression, Nc,Rd, for class 1 section is given by

N

Afc,Rd

y

M

.

= =×

γ 0

7580 2751 00

= 2084.5 × 103 N = 2084.5 kN > NEd = 1400 kN OK

BUCKLING RESISTANCE OF MEMBEREffective length of column about both axes is given by

L cr = L cr y = L cr z = 1.0L = 1.0 × 6000 = 6000 mm

The column will buckle about the weak (z–z) axis. Slenderness value to determine relative slenderness, λ1, is given by

λ π π1

3210 10275

86 8

.= =×

=Efy

Slenderness ratio about z–z axis (λ z) is

dz

y

cr

cr

z

.

.

.= = × = × =AfN

Li

1 600051 9

186 8

1 331λ

hb

..

. .= = <209 6205 2

1 02 1 2 and tf = 14.2 mm < 100 mm. Hence, from Table 9.12, for buckling about z–z axis buckling

curve c is appropriate and from Table 9.11, α = 0.49

Φ = 0.5[1 + α(d − 0.2) + d2] = 0.5[1 + 0.49 × (1.33 − 0.2) + 1.332] = 1.66

χ

. . . .=

+ −=

+ −=

1 11 66 1 66 1 33

0 3772 2 2Φ Φ d2

Hence design buckling resistance, Nb,Rd, is given by

N

Afb,Rd

y

M1

=χγ

=

× ×× = <−

.

. .

0 377 7580 275

1 0010 785 8 14003 kN kN

The section is therefore unsuitable to resist the design force.


Design of columns

9780415467193_C09 9/3/09, 4:30 PM409


410

Example 9.8 Analysis of a column with a tie-beam at mid-height (EC 3)Recalculate the axial compression resistance of the column in Example 9.7 if a tie-beam is introduced at mid-heightsuch that in-plane buckling about the z–z axis is prevented (see below).

RESISTANCE OF CROSS-SECTION – COMPRESSIONDesign resistance of section for uniform compression, Nc,Rd = 2084.5 kN, as above.

BUCKLING RESISTANCE OF MEMBERBuckling about y–y axisEffective length of column about y-y axis is given by

Leff y = 1.0L = 1.0 × 6000 = 6000 mmSlenderness value to determine relative slenderness, λ1, is given by

λ π π1

3210 10275

86 8

.= =×

=Efy

Slenderness ratio about y–y axis (λ y) is

dy

y

cr

cr

y

.

.

.= = × = × =AfN

Li

1 600089 6

186 8

0 771λ

hb

..

. .= = <209 6205 2

1 02 1 2 and tf = 14.2 mm < 100 mm. Hence, from Table 9.12, buckling curve b is appropriate and

from Table 9.11, α = 0.34Φ = 0.5[1 + α(d − 0.2) + d2]

= 0.5[1 + 0.34 × (0.77 − 0.2) + 0.772] = 0.893

χyχ

. . .

.=+ −

=+ −

=1 1

0 893 0 893 0 770 743

2 2 2Φ Φ d2

Hence design buckling resistance about the y–y axis is given by

N

Afb,Rd

y

M1

=χγ

=

× ×× = >−

.

. .

0 743 7580 275

1 0010 1548 8 14003 kN kN

Buckling about z–z axisEffective length of column about z–z axis, L cr z, is equal to 3000 mm

NEd

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411

Design of columns

NEd = 810 kN

MEd,b = 0

MEd,m = 225 kNm

MEd,t = 450 kNm

Slenderness ratio about z–z axis (λz) is

dz

y

cr

cr z

y

.

.

.= = × = × =AfN

Li

1 300051 9

186 8

0 6661λ

From above h/b = 1.02 < 1.2 and tf = 14.2 mm. Hence, from Table 9.12, buckling curve c is appropriate and, fromTable 9.11, α = 0.49

Φ = 0.5[1 + α(d − 0.2) + d2] = 0.5[1 + 0.49 × (0.666 − 0.2) + 0.6662] = 0.836

χz

. . .

.=+ −

=+ −

=1 1

0 836 0 836 0 6660 745

2 2 2Φ Φ d2

N

Afb,Rd

y

M1

kN kN .

. = =

× ×× = >−χ

γ0 745 7580 275

1 0010 1553 14003

Hence, compression resistance of column is 1548.8 kN > 1400 kN OK

Example 9.9 Analysis of a column resisting an axial load and moment(EC 3)

A 305 × 305 × 137 UC section extends through a height of 3.5 metres and is pinned at both ends. Check whether thismember is suitable to support a design axial permanent load of 600 kN together with a major axis variable bendingmoment of 300 kNm applied at the top of the element. Assume S275 steel is to be used and that all effective lengthfactors are unity.

DESIGN SUMMARYAxial permanent action = 600 kNAxial variable action = 0 kNPermanent bending action = 0 kNVariable bending action = 300 kNmGrade S275 steelEffective length factors for both axial and lateral torsional buckling are unity

ACTIONSFactored axial load is

NEd = 600 × 1.35 = 810 kN

Factored bending moments at top, middle and bottom of column are

MEd,t = 300 × 1.5 = 450 kNm

MEd,m = 225 kNm

MEd,b = 0 kNm


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412

Example 9.9 continuedSTRENGTH CLASSIFICATION(from Table 9.4)Flange thickness tf = 21.7 mm, so fy = 265 N mm−2

SECTION CLASSFICATION(From Table 9.5)

ε = (235/fy)0.5 = (235/265)0.5 = 0.94

ct f

..

=132 25

21 7 = 6.09 < 9ε = 9 × 0.94 = 8.46

(where c = (b − tw − 2r )/2 = (308.7 − 13.8 − 2 × 15.2)/2 = 264.5/2 = 132.25 mm)

Also

ct

dt

∗= =

w

.

.246 613 8

= 17.87 < 33ε = 33 × 0.94 = 31

Hence from Table 9.5, section belongs to class 1.

RESISTANCE OF CROSS-SECTIONS: BENDING AND AXIAL FORCESince cross-section is class 1 check

MEd ≤ MN,Rd

0 25 0 25 0 25

17500 265

1 0010 1159 4

0

3. . .

. . N

AfNpl,Rd

y

MEdkN= × = ×

×× = >−

γOK

0 5 0 5 320 5 2 21 7 13 8 265

1 0010 506 7

0

3.

kNw w y

MEd

h t fN

γ

. ( . . ) .

. . =

× − × × ×× = <−

Hence, allowance needs to be made for the effect of axial load on the plastic moment of resistance moment aboutthe y–y axis.

n

NN

NAf

( / )

( / . )

.= = =× ×

=−

Ed

pl,Rd

Ed

y Mγ 03

81017500 265 1 00 10

0 175

a

A btA

( )

( . . )

.=−

=− × ×

=2 17500 2 308 7 21 7

175000 234f

M

W fpl,y,Rd

pl,y y

M

kNm

. . = =

× ×× =−

γ 0

362300 10 265

1 0010 609 5

Since n < a

M M

n

aMN,y,Rd pl,y,Rd EdkNm kNm

( )

( . ) .

( . )

( . . ) . =

−−

=−

− ×= > =

1

1 0 5609 5

1 0 175

1 0 5 0 234569 5 450 OK

RESISTANCE OF MEMBER: COMBINED BENDING AND AXIAL COMPRESSION

For buckling about y–yEffective length of column about y–y, Lcr y, is given by

Lcr y = 1.0L = 1.0 × 3500 = 3500 mm

9780415467193_C09 9/3/09, 4:31 PM412

413

N

EI

Lcr,y

y

cr,y

N

= =× × × ×

=π π2

2

2 3 4

2

210 10 32800 10

350055495376

dy

y

cr,y

.= =× ×

=AfN

175 10 26555495376

0 292

hb

..

. .= = <320 5308 7

1 04 1 2 and tf = 21.7 mm. Hence from Table 9.12, for buckling about y–y axis buckling curve b is

appropriate and from Table 9.11, α = 0.34.

Φy = 0.5[1 + α(d y − 0.2) + d2y] = 0.5[1 + 0.34 × (0.29 − 0.2) + 0.292] = 0.56

χy

y y2 2

. . .

. .=+ −

=+ −

= <1 1

0 56 0 56 0 290 96 1 0

2 2Φ Φ dOK

For buckling about the minor axis, z–zEffective length of column about z–z, L cr,z, is given by

L cr z = 1.0L = 1.0 × 3500 = 3500 mm

N

EI

Lcr,z

z

cr,z

N

= =× × × ×

=π π2

2

2 3 4

2

210 10 10700 10

350018103674

dz

y

cr,z

.= =× ×

=AfN

175 10 26518103674

0 512

hb

..

. .= = <320 5308 7

1 04 1 2 and tf = 21.7 mm. From Table 9.12, for buckling about z–z axis use buckling buckling curve c.

Hence, from Table 9.11, α = 0.49.

Φz = 0.5[1 + α(dz − 0.2) + d2z] = 0.5[1 + 0.49 × (0.51 − 0.2) + 0.512] = 0.71

χz

z z2

z2

. . .

. .=+ −

=+ −

= <1 1

0 71 0 71 0 510 83 1 0

2 2Φ Φ dOK

MEMBER BUCKLING RESISTANCE IN BENDING

My,Rk

My,Rk = Wyfy = 2300 × 103 × 265 = 609.5 × 106 Nmm = 609.5 kNm

where Wy = Wpl,y (= 2300 cm3) for class 1 cross-sections

χLT

M

EIL

II

L GIEIcr

z

cr

w

z

cr t

z

.

= +

ππ

2

2

2

2

0 5

=

× × × × ××

+× × ×

× × × ×

= ×

( ).

.

. .

ππ

2 3 4

2

12

4

2 6

2 3 4

0 5

9210 10 10700 103500

2 38 1010700 10

3500 81000 2 5 10210 10 10700 10

3 31 10 Nmm

dLT

y y

cr

.

. .= =

××

=W fM

609 5 103 31 10

0 436

9


Design of columns

9780415467193_C09 9/3/09, 4:31 PM413


414

From above, h/b = 1.04 < 2 and tf = 21.7 mm. Hence, from Table 9.8, buckling curve a is appropriate ⇒ αLT = 0.21(Table 9.7 )

ΦLT = 0.5[1 + αLT(d LT − 0.2) + d2LT]

= 0.5[1 + 0.21(0.43 − 0.2) + 0.432] = 0.62

χLT

LT LT2

LT2

=+ −

1

Φ Φ d

=

+ −= <

. . . . .

10 62 0 62 0 43

0 94 1 02 2

OK

Mz,Rk

Mz,Rk = Wzfy = Wpl,zfy = 1050 × 103 × 265 = 278.25 × 106 Nmm = 278.25 kNm

Equivalent uniform moment factor, Cmi (Table 9.13)Cmi = 0.6 + 0.4ψ ≥ 0.4

ψ = 0 (for bending about y–y)

⇒ Cmy = 0.6

Considering bending about y–y and out of plane supports

ψ = 0 (for bending about y–y)

⇒ CmLT = 0.6

kyy, kzyFor class 1 cross-sections

k C

NN

CN

Nyy my yEd

y Rk Mmy

Ed

y Rk M

[ . ]/

./

= + −

≤ +

1 0 2 1 0 8

1 1

dχ γ χ γ

= + −

×× ×

= . [ . . ]

. . / . .0 6 1 0 29 0 2

810 100 96 4637 5 10 1 00

0 613

3

≤ +

= +

×× ×

= ./

. .

. . / . .C

NNmy

Ed

y Rk M

1 0 8 0 6 1 0 8810 10

0 96 4637 5 10 1 000 69

1

3

3χ γ

Hence k yy = 0.61

k

CN

N CN

Nzyz

mLT

Ed

z Rk M mLT

Ed

z Rk M

. .

/

.

.

/= −

−× ≥ −

−×1

0 10 25

10 1

0 251 1

dχ γ χ γ

= −

×−

××

× ×=

. .. .

. . / . .1

0 1 0 510 6 0 25

810 100 83 4637 5 10 1 00

0 973

3

≥ −

−× = −

−×

×× ×

= .

.

/

.. .

. . / . .1

0 10 25

10 1

0 6 0 25810 10

0 84 4637 5 10 1 000 94

1

3

3CN

NmLT

Ed

z Rk Mχ γ

Hence k zy = 0.97


9780415467193_C09 9/3/09, 4:32 PM414

415

Design of columns

SECTION SELECTIONAs before, try a 203 × 203 × 52 UC.

DESIGN LOADINGS AND MOMENTSUltimate reaction from beam A, RA = 200 kN; ultimate reaction from beam B, RB = 75 kN; assume self-weight ofcolumn = 5 kN. Ultimate axial load, NEd, is

NEd = RA + RB + self-weight of column

= 200 + 75 + 5 = 280 kN

Ultimatereaction fromBeam B – 75 kN

Ultimatereaction fromBeam A – 200 kN

Plan on capL = 7 m

F

Ultimate load dueto self-weight 5 kN

LE = 0.85L

Effectively held in positionbut not restrained indirection, i.e. pinned

Effectively held in positionand direction, i.e. fixed

Interaction equations

NN

kM

Mk

MM

Ed

y Rk Myy

y,Ed

LT y,Rk Myz

z,Rd


/

/1 1 1

+ +

=

×× ×

+×

× ×+

. . / .

.

. . / .

810 100 96 4637 5 10 1 00

0 61450 10

0 94 609 5 10 1 000

3

3

6

6

= 0.18 + 0.48 = 0.66 < 1.0 OK

NN

kM

Mk

MM

Ed

z Rk Mzy

y,Ed

LT y,Rk Mzz

z,Rd


/

/1 1 1

+ +

=

×× ×

+×

× ×+

. . / .

.

. . / .

810 100 83 4637 5 10 1 00

0 97450 10

0 94 609 5 10 1 000

3

3

6

6

= 0.21 + 0.76 = 0.97 < 1.0 OK

Hence the selected section is suitable.

Example 9.10 Analysis of a steel column in ‘simple’ construction (EC 3)Check the column in Example 4.12 using the simplified approach discussed in 9.12.3.


9780415467193_C09 9/3/09, 4:32 PM415


416

Load eccentricity for beam A,

ey = h/2 + 100 = 206.2/2 + 100 = 203.1 mm

Load eccentricity for beam B,

ez = tw/2 + 100 = 8/2 + 100 = 104 mm

Moment due to beam A,

My = RAey = 200 × 103 × 203.1 = 40.62 × 106 Nmm

Moment due to beam B,

Mz = RBez = 75 × 103 × 104 = 7.8 × 106 Nmm

AXIAL BUCKLING RESISTANCEBy inspection, buckling about z–z will determine the compression strength of the column.

tf = 12.5 mm ⇒ fy = 275 Nmm−2 (Table 9.4)

ε .= = =

235 235275

0 924fy

λ π ε1

y

. . . .= = = × =Ef

93 9 93 9 0 924 86 8

L cr,z = 0.85L = 0.85 × 7000 = 5950 mm (Table 4.15)

λz

cr,z

z

.

.= = =Li

595051 6

115 3

dz

z ..

.= = =λλ1

115 386 8

1 33

hb

..

. .= = <206 2203 9

1 01 1 2

Hence from Table 9.12 for buckling about z–z use buckling curve c and from Table 9.11 α = 0.49.

Φ = 0.5[1 + α(d − 0.2) + d2] = 0.5[1 + 0.49(1.33 − 0.2) + 1.332] = 1.66

χz 2

. . .

. .=+ −

=+ −

= ≤1 1

1 66 1 66 1 330 38 1 0

2 2 2Φ Φ dOK

N

Afb,z,Rd

z y

M

kN . .

. . = =

× × ×× =−χ

γ 1

230 38 66 4 10 275

1 010 693 9

LATERAL TORSIONAL BUCKLING RESISTANCE Mb,Rd

Length of column between points which are laterally restrained, L cr = 5950 mm.

M

EIL

II

L GIEIcr

z

cr

w

z

cr t

z

.

= +

ππ

2

2

2

2

0 5

=

× × × × ××

+× × ×

× × × ×

= ×

.

ππ

2 3 4

2

9

4

2 3

2 3 4

0 5

6210 10 1770 105950

166 101770 10

5950 81000 320 10210 10 1770 10

192 10 Nmm


d LT

y y

cr

.= =

× ××

=W fM

568 10 275192 10

0 903

7


9780415467193_C09 9/3/09, 4:33 PM416

417

From above h/b = 1.01 < 2. From Table 9.9 use buckling curve b ⇒ αLT = 0.34 (Table 9.7 )

ΦLT = 0.5[1 + αLT(dLT − dLT,0) + βd2LT] = 0.5[1 + 0.34(0.9 − 0.4) + 0.75 × 0.92] = 0.89

χ

βLTLT LT

2LT2

LT2

. . .

. . .

.=+ −

=+ − ×

= < < = =1 1

0 89 0 89 0 75 090 76 1 0

1 10 9

1 22 2 2Φ Φ d d

OK

M

W fb,Rd

LT y y

M

.

. = =

× × ×× =−χ

γ 1

360 76 568 10 275

1 010 118.7 kNm

BENDING RESISTANCE Mz,Rd

M

W fz,Rd

pl,z y

M

. = =

× ×× =−

γ 1

36174 10 275

1 010 47.85 kNm

Interaction equation

NN

MM

MM

Ed

b,z,Rd

y,Ed

b,Rd

z,Ed

z,Rd

. + + ≤1 5 1

280693 9

40 4118 7

1 57 8

47 850 40 0 34 0 245 0 985 1

.

..

. ..

. . . . + + × = + + = < OK

Example 9.11 Analysis of a column baseplate (EC 3)Check that the column baseplate shown below is suitable to resist an axial design load, NEd, of 2200 kN. Assume thatthe foundations are of concrete of compressive cylinder strength, fck, of 30 N mm−2 and that the baseplate is made ofS275 steel.

Design of columns


EFFECTIVE AREASince the plate is between 16–40 mm thick and is made of S275 steel, fy = 265 Nmm−2 (Table 9.4)Additional bearing width, x, is

NEd = 2200 kN

9780415467193_C09 9/3/09, 4:33 PM417


418


x t

ff

. .

/ /

=

=

× ×

y

j M330

2653 13 3 1 00

1 2 1 2

γ= 77 mm < 0.5(500 − 314.5) = 92 mm OK

where

fj = βjk jfcd = 2/3 × 1.0 × (1.00 × 30/1.5) = 13.3 Nmm−2

Aeff = (2x + h)(2x + b) − (b − tw)(h − 2tf − 2x)

= (2 × 77 + 314.5)(2 × 77 + 306.8) − (306.8 − 11.9)(314.5 − 2 × 18.7 − 2 × 77)

= 215885 − 36302 = 179583 mm2

AXIAL LOAD CAPACITYAxial load capacity of baseplate = Aefff j

= 179583 × 13.3 × 10−3

= 2388 kN > NEd = 2200 kN OK

BENDING IN BASEPLATEBending moment per unit length in baseplate, mEd, is

mEd = (x2/2)NEd /Aeff = (772/2)2200/179583 = 36.3 kNmm mm−1

Moment of resistance, mRd is

m

t fRd

y

M

.

= =××

2

0

3

630 265

6 1 0γ = 39750 N mm mm−1 = 39.75 kN mm mm−1 > mEd OK

9.13.1 MATERIAL PROPERTIES

9.13.1.1 Nominal bolt strengths (Cl. 3.3,EC 3–8)

The recommended bolt classes and associatednominal values of the yield strength fyb and theultimate tensile strength fub (to be adopted as char-acteristic values in design calculations) are shownin Table 9.14.

Cl. 3.1.2 of EC 3–8 recommends that only boltclasses 8.8 and 10.9 conforming to the require-ments given in EN 14399: Group 4: High strengthstructural bolting for preloading, with controlled tight-ening in accordance with EN1090–2: Group 7:Execution of steel structures may be used as preloadedbolts.

9.13 ConnectionsConnection design is covered in Part 1.8 ofEurocode 3 (EC 3–8). The guidance provided ismore comprehensive than BS 5950, but the resultsseem broadly similar and the principles are essen-tially the same. One exception may be the designof friction grip fasteners, where the slip factor foruntreated surfaces may have to be taken as 0.2rather than 0.45 in BS 5950. In general the resultsfor bolting and welding seem slightly more con-servative than BS 5950. This is largely becauseof the larger partial safety factors for connectionsγM = 1.25.

To help comparison of the design methods inBS 5950 and EC 3 with regard to connections,the material in this section is presented under thefollowing headings:

1. material properties2. clearances in holes for fasteners3. positioning of holes for bolts4. bolted connections5. high strength bolts in slip-resistant connections6. welded connections7. design of connections.

Table 9.14 Nominal values of fyb and fub forbolts (Table 3.1, EC 3–8)

Bolt classes 4.6 4.8 5.6 5.8 6.8 8.8 10.9

fyb (N mm−2) 240 320 300 400 480 640 900fub (N mm−2) 400 400 500 500 600 800 1000

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419

9.13.1.2 Filler metalsClause 4.2 (2) of EC 3–8 recommends that thespecified yield strength, ultimate tensile strength,etc., of the filler metals used in welded connec-tions should be equal to or greater than the corre-sponding values specified for the steel being welded.

9.13.1.3 Partial safety factor (Cl. 2.2, EC 3–8)Partial safety factors, γM, are given in Table 2.1of EC 3–8. Those relevant to this discussion arereproduced below

Resistance of bolts and welds γM2 = 1.25Slip resistance at ULS (Cat C) γM3 = 1.25Slip resistance at SLS (Cat B) γM3,ser = 1.25

Note that the above values appear in the normativeversion of EC 3–8 and the accompanying NationalAnnex may specify other values.

9.13.2 CLEARANCES IN HOLES FOR FASTENERS(SEE REFERENCE STANDARDS:GROUP 7)

The nominal clearance in standard holes for boltedconnections should be as follows:

(i) 1 mm for M12 and M14 bolts(ii) 2 mm for M16 and M24 bolts(iii) 3 mm for M27 and larger bolts.

The nominal clearance in oversize holes for slip-resistant connections should be:

(i) 3 mm for M12 bolts(ii) 4 mm for M14 to M22 bolts(iii) 6 mm for M24 bolts(iv) 8 mm for M27 and larger bolts.

9.13.3 POSITIONING OF HOLES FOR BOLTS(CL. 3.5, EC 3–8)

9.13.3.1 Minimum end and edge distancesThe end distance e1 from the centre of a fastenerhole to the adjacent end of any part, measured inthe direction of load transfer (see Fig. 9.8), shouldbe not less than 1.2do, where do is the hole diameter.

Similarly the edge distance e2 from the centre ofa fastener hole to the adjacent edge of any part,measured at right angles to the direction of loadtransfer (Fig. 9.8), should not be less than 1.2do.

9.13.3.2 Maximum end and edge distancesUnder normal conditions, the end and edge dis-tance should not exceed 8t or 125 mm, whicheveris the larger, where t is the thickness of the thinnerouter connected part.

9.13.3.3 Minimum and maximum spacingbetween fasteners

The spacing p1 between centres of fasteners in thedirection of load transfer (Fig. 9.8), should be notless than 2.2do nor generally exceed the smaller of14t and 200 mm.

The spacing p2 between rows of fasteners, meas-ured perpendicular to the direction of load transfer(Fig. 9.8), should not be less than 2.4do nor gener-ally exceed the smaller of 14t and 200 mm.

9.13.4 BOLTED CONNECTIONS (CL. 3.6, EC 3–8)

9.13.4.1 Design shear resistance per shearplane

If the shear plane passes through the threadedportion of the bolt, the design shear resistance pershear plane, Fv,Rd, is given by:

F f

Av,Rd v ub

s

M

= αγ 2

(9.78)

If the shear plane passes through the unthreadedportion of the bolt, the design shear resistance isgiven by:

F f

Av,Rd ub

M

.= 0 62γ

(9.79)

whereA is the gross cross-section of the boltAs is the tensile stress area of the bolt

(Table 4.22)fub is the ultimate tensile strength of the bolt

(Table 9.14)α v is a factor obtained from Table 9.15

Table 9.15 Values for α v

Bolt classes 4.6 4.8 5.6 5.8 6.8 8.8 10.9

αv 0.6 0.5 0.6 0.5 0.5 0.6 0.5

Connections

Fig. 9.8 Spacing of fasteners (based on Table 3.3,EC 3–8).

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420

µ = 0.5 for class A surfacesµ = 0.4 for class B surfacesµ = 0.3 for class C surfacesµ = 0.2 for class D surfaces

whereClass A are surfaces blasted with shot or grit,

with any loose rust removed, no pitting;surfaces blasted with shot or grit, and spray-metallised with aluminium; surface blastedwith shot or grit, and spray-metallised with azinc-based coating certified to provide a slipfactor of not less than 0.5.

Class B are surfaces blasted with shot or grit,and painted with an alkali-zinc silicate paintto produce a coating thickness of 50–80 µm.

Class C are surfaces cleaned by wire brushing orflame cleaning, with any loose rust removed;

Class D are surfaces not treated. (Table 3.7,EC 3–8)

9.13.6 WELDED CONNECTIONS (CL. 4.5.3.3,EC 3–8)

9.13.6.1 Design resistance of a fillet weldEC 3–8 recommends two methods for the designof fillet welds namely the ‘directional method’ andthe ‘simplified method’. The latter is similar tothat used in BS 5950 and is the only one discussedhere. According to the simplified method the designresistance of a fillet weld will be adequate if at allpoints the design value of the weld force per unitlength, Fw,Ed, is less than or equal to the designweld resistance per unit length, Fw,Rd i.e.

Fw,Ed ≤ Fw,Rd (9.87)

The design weld resistance per unit length of afillet weld is given by

Fw,Rd = fvw,da (9.88)

wherea is the throat thickness of the weld and

is taken as the height of the largesttriangle which can be inscribed within thefusion faces and weld surface, measuredperpendicular to the outer side of thistriangle (Fig. 9.9). Note that a shouldnot be less than 3mm.

fvw,d is the design shear strength of the weld andis given by:

f

fvw,d

u

w M

/

=3

2β γ(9.89)

It should be noted that these values for designshear resistance apply only where the bolts areused in holes with nominal clearances specified insection 9.13.2.

9.13.4.2 Bearing resistanceThe design bearing resistance, Fb,Rd, is given by

F k f

dtbb,Rd 1 u

M

= αγ 2

(9.80)

where

k1 = min

2 8 1 7 2 52. . ; .edo

−

for edge bolts (9.81)

k1 = min 1 4 1 7 2 52. . ; .

pdo

−

for inner bolts (9.82)

αb = min

ed

ff

1

31 0

o

ub

u

; ; .

for end bolts (9.83)

αb = min

pd

ff

1

314

1 0o

ub

u

; ; .−

for inner bolts

(9.84)

9.13.5 HIGH-STRENGTH BOLTS INSLIP-RESISTANT CONNECTIONS(CL. 3.9, EC 3–8)

9.13.5.1 Slip resistanceThe design slip resistance of a preloaded high-strength bolt (i.e. bolt classes 8.8 and 10.9), Fs,Rd,is given by

F

k n Fs,Rd

s p,C

M

=µ

γ 3

(9.85)

whereks = 1.0 where the holes in all the plies have

standard nominal clearances as outlined insection 9.13.2 above

n is the number of friction interfacesµ is the slip factor (see below)γM3 is the partial safety factor. For bolt holes

in standard nominal clearance holes,γM3 = 1.25 and 1.10 for the ultimateand serviceability limit states respectively

Fp,C is the design preloading force. It is given by:

Fp,C = 0.7fubA s (9.86)

9.13.5.2 Slip factorThe value of the slip factor µ is dependent on theclass of surface treatment. According to Table 3.7of EC 3–8 the value of µ should be taken as follows

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421

Fig. 9.10 Splice connection.

9.13.7.1 Splice connectionsThe design of splice connections (Fig. 9.10) inEC 3 is essentially the same as that used in BS5950 and involves determining the design values ofthe following parameters:

(1) design shear resistance of fasteners (section9.13.4 or 9.13.5)

(2) critical bearing resistance (section 9.13.4)(3) critical tensile resistance (see below).

Tension resistance of cross-sections (Cl. 6.2.3,EC 3–1). For members in axial tension, thedesign value of the tensile force NEd at eachcross-section should satisfy the following

NN

Ed

t,Rd

.≤ 1 0 (9.90)

whereNt,Rd is the design tension resistance of the

cross-section.

For sections with holes the design tension resist-ance should be taken as the smaller of:

N

Afpl,Rd

y

M

=γ 0

(9.91)

N

A fu,Rd

net u

M2

.

=0 9

γ(9.92)

whereNpl,Rd is the design plastic resistance of the gross

cross-sectionNu,Rd is the design ultimate resistance of the net

cross-section at holes for fasteners

In Category C connections, i.e. slip-resistant atultimate limit state, the design tension resistanceof the cross-section should be taken as Nnet,Rd, givenby

N

A fnet,Rd

net y

M

=γ 0

(9.93)

Connections

Fig. 9.9 Throat thickness of a fillet weld.

wherefu is the nominal ultimate tensile strength of

the weaker part joined andβw is a correlation factor whose values should

be taken from Table 9.16.

9.13.7 DESIGN OF CONNECTIONSThe use of the above equations is illustrated bymeans of the following design examples:

1. tension splice connection;2. welded end plate and beam connection;3. bolted beam-to-column connection using end

plates;4. bolted beam-to-column connection using web

cleats.

Table 9.16 Ultimate tensile strength andcorrelation factor βw for fillet welds (based onTable 4.1, EC 3–8)

EN10025 Ultimate tensile Correlationsteel grade strength fu (N mm−2) factor βw

S235 360 0.8S275 430 0.85S355 510 0.9

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422

Example 9.12 Analysis of a tension splice connection (EC 3)Calculate the design resistance of the connection detail shown below. The cover plates are made of S275 steel andconnected with either

a) non-preloaded bolts of diameter 20 mm and class 4.6 orb) prestressed bolts of diameter 16 mm and class 8.8

Assume that in both cases, the shear plane passes through the unthreaded portions of the bolts.

NON-PRELOADED BOLTS

Design shear resistanceDesign shear resistance per shear plane, Fv,Rd, is given by

F f

Av,Rd v ub

s

M

N kN . .

= = × × = =αγ 2

0 6 400314

1 2560318 60

where

α v = 0.6 for class 4.6 bolts

fub = 400 N mm−2 (Table 9.14)

γMb = 1.25

A

d

mm= =

×=

π π2 22

4

20

4314

All four bolts are in double shear. Hence, shear resistance, FEd, of connection is

FEd = 4 × (2 × 60) = 480 kN

Bearing resistanceBearing failure will tend to take place in the cover plates since they are thinner. According to Table 3.4 of EC 3–8,αb is the smallest of

For end bolts, αb = min

ed

ff

1

335

3 220 53

400430

0 93 1 0o

ub

u

. ; . ; .=×

= = =

For inner bolts, αb = min

pd

ff

1

314

703 22

14

0 81 0 93 1 0o

ub

u

. ; . ; .− =×

− = =

Hence, αb may be conservatively taken as 0.53.

For edge bolts, k1 = min

2 8 1 7 2 83522

4 45 2 52. . . . ; .edo

− = × =

For inner bolts, k1 = min

1 4 1 7 1 47022

4 45 2 52. . . . ; .pdo

− = × =

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9.13.7.2 Welded end-plate to beamconnection

The relevant equations for the design of this typeof connection were discussed in section 9.13.6.

Connections

Hence, k1 = 2.5 and the design bearing resistance of one shear plane, Fb,Rd, is given by

F k f

dtb,Rd b u

M

N kN . .

. . = = × × ×

×= =1

2

2 5 0 53 43020 6

1 2554696 54 7α

γBearing resistance of double shear plane is

2 × 54.7 = 109.4 kN

Bearing resistance of bolt group is

4 × 109.4 = 437.6 kN

Tensile resistance of cover platesThe design tension resistance of the cover plates, Nt,Rd, should be taken as the smaller of the design plastic resistance ofthe gross cross-section, Npl,Rd, and the design ultimate resistance of the net cross-section at holes for fasteners, Nu,Rd:

N

Afpl,Rd

y

M

( )

. = =

× ×× =−

γ 0

36 140 2751 0

10 231 kN

Net area of cover plate, Anet = 6 × 140 − 2 × 6 × 22 = 576 mm2

N

A fu,Rd

net u

M

kN .

.

. . = =

× ×× =−0 9 0 9 576 430

1 1010 202 6

2

3

γ (critical)

Total ultimate resistance of connection i.e. two cover plates = (2 × 202.6) = 405.2 kN. Hence design resistance of theconnection with class 4.6, 20 mm diameter non-preloaded bolts is 405 kN.

PRESTRESSED BOLTSSlip resistancePreloading force, Fp,C, is given by

Fp,C = 0.7fubA s = 0.7 × 800 × 157 = 87920 N = 87.9 kN

Assuming the surfaces have been shot blasted, i.e. class A, take µ = 0.5. For two surfaces n = 2, standard clearancesk s = 1.0 and γM3 = 1.25Hence, slip resistance for each bolt, Fs,Rd, is given by:

F

k n Fs,Rd

s p,C

M

kN . .

. . = =

× × ×=

µγ 3

1 2 0 5 87 9

1 2570 3

Hence design resistance = 4 × 70.3 = 281.2 kN

Tensile resistance of cover platesNet area of cover plate, Anet = 6 × 140 − 2 × 6 × 18 = 624 mm2

Tensile resistance of cover plate is given by

N

A fnet,Rd

net y

M

kN

. . = =

×× =−

γ 0

3624 275

1 010 171 6

Total ultimate resistance of connection i.e. two cover plates = (2 × 171.6) = 343.2 kN. Hence, design resistance ofconnection made with grade 8.8, 16 mm diameter prestressed bolts is 281 kN.

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Example 9.13 Shear resistance of a welded end plate to beamconnection (EC 3)

Calculate the shear resistance of the welded end plate-to-beam connection shown below. Assume the throatthickness of the fillet weld is 4 mm and the steel grade is S 275.

9.13.7.3 Bolted beam-to-column connectionusing an end plate

The design of this type of connection (Fig. 9.11)involves carrying out the following checks:

1. ductility of the material/joint (see below)2. shear resistance of fasteners (see 9.13.4)3. bearing resistance of fasteners (see 9.13.4)4. resistance of welded connections (see 9.13.6)5. shear resistance of end plate (see below)6. local shear resistance of beam web (see 9.11.1.2)7. block tearing failure (see below)

Ductility requirements (Cl. 3.2.2, EC 3 &NA.2.5). EC 3 requires that a check is carriedout on the ductility of the materials being joined.According to clause 3.2.2(2) the ductility require-ment is generally achieved provided that the ratioof the specified minimum ultimate tensile strength,fu, to the specified minimum yield strength, fy,exceeds 1.1, i.e.

Fig. 9.11 Typical bolted beam-to-column connection.

ffu

y

.≥ 1 1 (9.94)

In the case of simple end plate connections,document SN014a (SCI) recommends that thefollowing should also be satisfied:

hp = 240 mm

Ultimate tensile strength of S 275 steel (fu) = 430 N mm−2 (Table 9.16)Correlation factor (βw) = 0.85Throat thickness of weld (a) = 4 mmPartial safety factor for welds (γM2) = 1.25Design shear strength of weld, fvw,d, is given by

f

fvw,d

u

w M

N mm /

/

. . . = =

×= −3 430 3

0 85 1 25233 6

2

2

β γ

Design resistance of weld per unit length, Fw,Rd, is given by

Fw,Rd = fvw,da = 233.6 × 4 = 934.4 N mm−1

Weld length, L = 240 − 2a = 240 − 2 × 4 = 232 mm and hence the shear resistance of the weld, Vw,Rd, is

Vw,Rd = 2 × (Fw,Rd × L) = 2 × (934.4 × 232) = 433561 N = 433 kN

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425

NEd NEd

NEd

NEd

small tension force

large shear force

small shear force

large tension force

Fig. 9.12 Block tearing – net tension and shear areas (Fig. 3.8, EC 3–8).

If the supporting element is a beam or colum web

t

d ffp

ub

y,plate

.

≤2 8

(9.95)

If the supporting element is a column flange

t

d ff

td f

fpub

y,platef,c

ub

y,column

or .

.

≤ ≤2 8 2 8

(9.96)

Shear resistance of end plate (Cl. 6.2.6, EC 3).The design value of the shear force VEd at eachcross section should satisfy the following

VV

Ed

c,Rd

≤ 1 (9.97)

where Vc,Rd is the design shear resistance.For plastic design Vc,Rd is equal to the design

plastic shear resistance, Vpl,Rd, given by

V

A fpl,Rd

v y

M

( / )

=3

0γ(9.98)

whereAv is the (gross) shear area and according to

document SN014a (Steel ConstructionInstitute) may be taken as

A

h tv

p p .

=1 27

(9.99)

The coefficient 1.27 takes into account the reduc-tion of the shear resistance, due to the presence ofin plane bending moments.

Where fastener holes are present the design shearresistance is equal to VRd,net, given by

V

A fRd,net

v,net u

M

( / )

=3

2γ(9.100)

wherefu is the ultimate tensile strength of the plate

material (Table 9.14)Av,net is the net cross-section area of the plate

= tp(hp − nd0) (9.101)

in whichtp is the thickness of the platehp is the length of the platen is the number of bolt rowsd0 is the diameter of the bolt holes

Design for block tearing (Cl. 3.10.2, EC 3–8).‘Block tearing’ failure at a group of fastener holesnear the end of the beam web may occur as shownin Fig. 9.12. For a symmetric bolt group subject toconcentric loading the design block tearing resist-ance, Veff,1,Rd, is given by

V

f A f Aeff,1,Rd

u nt

M

y nv

M

( / )

= +γ γ2 0

3(9.102)

whereAnt is the net area subject to tensionAnv is the net area subject to shear

Connections

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426

Example 9.14 Bolted beam-to-column connection using an end plate(EC 3)

If the beam in Example 9.13 is to be connected to a column using eight class 8.8, M20 bolts as shown below,calculate the maximum shear resistance of the connection.

CHECK POSITIONING OF HOLES FOR BOLTS

Diameter of bolt, d = 20 mm

Diameter of bolt hole, do = 22 mm

End distance, e1 = 30 mm

Edge distance, e2 = 35 mm

Pitch, i.e. spacing between centres of bolts in the direction of load transfer, p1 = 60 mm

Gauge, i.e. spacing between columns of bolts, p2 = 90 mm

Thickness of end plate, tp = 10 mm


End and edge distances, e1 = 30 and e2 = 35 ≥ 1.2do = 1.2 × 22 = 26.4 mm OK

e1 and e2 ≤ larger of 8t (= 8 × 10 = 80 mm) or 125 mm > 30, 35 OK

Spacing, p1 ≥ 2.2do = 2.2 × 22 = 48.4 < 60 mm OK

Spacing, p2 ≥ 2.4do = 2.4 × 22 = 52.8 < 90 mm OK

Spacing, p1 and p2 ≤ lesser of 14t (= 14 × 10 = 140 mm) or 200 mm > 60, 90 OK

Class 8.8, M20 bolts

90VEd

2

VEd

2

hp

35 35

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DUCTILITY REQUIREMENTSSince the supporting element is a column flange, the following conditions apply:

t

d ff

td f

fpub

y,platef,c

ub

y,column

or≤ ≤ .

.2 8 2 8

d f

ft

2 8

20

2 8

800

27512 2 10

.

. . ub

y,platepmm mm= = > = OK

BEARING RESISTANCE OF BOLT GROUPSince bolt diameter, d = 20mm, hole diameter, do = 22mm and αb is the smallest of


ed

ff

1

330

3 220 455

800410

1 86 1 0o

ub

u,p

. ; . ; .=×

= = =


pd

ff

1

314

603 22

14

0 659 1 86 1 0o

ub

u

. ; . ; .− =×

− = =



2 8 1 7 2 83522

4 45 2 52. . . . ; .edo

− = × =


1 4 1 7 1 49022

5 7 2 52. . . . ; .pdo

− = × =

Hence, k1 = 2.5.The design bearing resistance of one shear plane, Fb,Rd, is given by

F k f

dtb,Rd b u

M

N kN . .

. . = = × × ×

×= =1

2

2 5 0 455 41020 10

1 2574620 74 6α

γBearing resistance of bolt group = 8Fb,Rd = 8 × 74.6 = 596.8 kN

SHEAR RESISTANCE OF BOLT GROUPDesign shear resistance per shear plane, Fv,Rd, is given by

F f

Av,Rd v ub

s

M

N kN . .

. = = × × = =αγ 2

0 6 800314

1 25120576 120 5

where

αv = 0.6 for class 8.8 bolts

fub = 800 N mm−2 (Table 9.14)

γM2 = 1.25

A

ds mm

= =

×=

π π2 22

4

20

4314

Hence, shear resistance of bolt group is

Vv,Rd = 8 × Fv,Rd = 8 × 120.5 = 964 kN

RESISTANCE OF WELDED CONNECTION BETWEEN BEAM AND END-PLATE

Vw,Rd = 433 kN (see Example 9.13)


Connections

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428

SHEAR RESISTANCE OF END PLATE (GROSS SECTION)

A

h tv

p p mm .

. . = =

×=

1 27

10 240

1 271889 8 2

Hence, the design plastic shear resistance of end plate, Vpl,Rd, per section is given by

V

A fpl,Rd

v y

M

kN ( / )

. ( / )

. = =

×× =−3 1889 8 275 3

1 0010 300

0

3

γFor failure, two planes have to shear. Hence, shear resistance of end plate is

2 × Vpl,Rd = 2 × 300 = 600 kN

SHEAR RESISTANCE OF END PLATE (NET SECTION)A v,net = tp(hp − nd0) = 10 × (240 − 4 × 22) = 1520 mm2

V

A fRd,net

v,net u

M

kN ( / )

( / )

. = = × =−3 1520 410 3

1 1010 327

2

3

γFor failure, two planes have to shear. Hence, shear resistance of end plate is

2 × VRd,net = 2 × 327 = 654 kN

LOCAL SHEAR RESISTANCE OF BEAM WEBBased on case (c) clause 6.2.6 of EC 3, shear area of web (A v)web is given by

(A v)web = 0.9Ltweb = 0.9 × 240 × 9.1 = 1965.6 mm2

where tweb is the beam web thickness = 9.1mmHence, local shear resistance of web, Vpl,Rd, is given by:

V

A fpl,Rd

y

M

kN ( / )

. ( / )

. = =

×× =−ν

γ3 1965 6 275 3

1 0010 312

0

3

BLOCK TEARINGAnt = (p2 − do)tp = (90 − 22) × 10 = 680 mm2 (Fig. 9.12)

Anv = 2(hp − e1 − 3.5do)tp = 2 × (240 − 30 − 3.5 × 22) × 10 = 2660 mm2

V

f A f Aeff,1,Rd

u nt

M

y n v

M

kN ( / )

.

( / )

. . = + =

×+

× =−

γ γ2 0

33 410 680

1 10

275 3 2660

1 0010 675 8

By inspection, the maximum shear resistance of the connection is controlled by the beam web in shear and is equalto 312 kN.


9.13.7.4 Bolted double angle web cleatbeam-to-column connection

The design of such connections (Fig. 9.13) involvesdetermining the design values of the followingparameters:

1. design shear resistance of fasteners (see 9.13.4)2. bearing resistance of fasteners (see 9.13.4)3. shear resistance of cleats (see previous case)

4. distribution of shear forces between fasteners(see below)

5. bearing resistance of beam web (see 9.13.4)6. block tearing failure (see previous case).

Distribution of forces between fasteners. Thedistribution of internal forces between fastenersat the ultimate limit state can be assumed to beproportional to the distance from the centre of

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429

10

35 45

rotation (see Fig. 9.14) where, amongst othercases, the design shear resistance Fv,Rd of a fasteneris less than the design bearing resistance Fb,Rd, i.e.

Fv,Rd < Fb,Rd (9.103)

Thus, for the connection detail shown in Fig.9.14, the horizontal shear force on the bolts, Fh,Ed,is given by:

F

Mph,EdEd =

5(9.104)

and the vertical shear force per bolts is = VEd/5.The design shear force, Fv,Ed, is given by

Fig. 9.13 Typical bolted double angle web cleat beam-to-column connection.

FM

pV

v,EdEd Ed

/

=

+

5 5

2 21 2

(9.105)

whereMEd = design bending momentVEd = design shear forcep = spacing between fasteners

Example 9.15 Bolted beam-to-column connection using web cleats(EC 3)

Show that the double angle web cleat beam-to-column connection detail shown below is suitable to resist the designshear force, VEd, of 200 kN. Assume the steel grade is S 275 and the bolts are class 8.8 and diameter 16 mm.

Connections

Fig. 9.14 Distribution of loads between fasteners(Fig. 6.5.7, ENV EC 3).

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430

CHECK POSITIONING OF HOLES FOR BOLTSDiameter of bolt, d = 16 mm

Diameter of bolt hole, do = 18 mm

End distance, e1 = 30 mm

Edge distance, e2 = 45 mm

Pitch i.e. spacing between centres of bolts in the direction of load transfer, p1 = 50 mm

Thickness of angle cleat, tc = 10 mm


End and edge distances, e1 = 30 and e2 = 45 ≥ 1.2do = 1.2 × 18 = 21.6 mm OK

e1 and e2 ≤ larger of 8t (= 8 × 10 = 80 mm) or 125 mm > 30, 45 OK

Pitch, p1 ≥ 2.2do = 2.2 × 18 = 39.6 < 50 mm OK

Pitch, p1 ≤ lesser of 14t (= 14 × 10 = 140 mm) or 200 mm > 50 OK

SHEAR RESISTANCE OF BOLT GROUP CONNECTING CLEATS TO SUPPORTING COLUMNAssume that the shear plane passes through threaded portion of the bolt. Hence, tensile stress area of bolt A s =157mm2 (Table 4.22). Shear resistance per bolt, Fv,Rd, is

F f

Av,Rd v ub

s

M

kN . .

. = = × × × =−αγ 2

30 6 800157

1 2510 60 3

where

α v = 0.6 for class 8.8 bolts

fub = 800 N mm−2 (Table 9.14)

γM2 = 1.25

A s = 157 mm2 (Table 4.22)

Hence, shear resistance of bolt group is

10 × Fv,Rd = 10 × 60.3 = 603 kN > VEd = 200 kN OK

BEARING RESISTANCE OF CLEATS CONNECTED TO SUPPORTING COLUMNBearing failure will tend to take place in the cleats since they are thinner than the column flange. Diameter of bolts(d ) 16 mm; hole diameter (do) = 18mm; end distance in the direction of load transfer (e1) = 30 mm; pitch (p1) =50 mm; ultimate tensile strength of class 8.8 bolts (fub) = 800 Nmm−2 (Table 9.14); ultimate tensile strength of S275steel, t ≤ 16 mm (fu) = 410 Nmm−2 (Table 9.4)Here αb is the smallest of


ed

ff

1

330

3 180 555

800410

1 95 1 0o

ub

u,c

. ; . ; .=×

= = =


pd

ff

1

314

503 18

14

0 676 1 95 1 0o

ub

u,c

. ; . ; .− =×

− = =



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431


2 8 1 7 2 84518

7 2 52. . . ; .edo

− = × =


1 4 1 7 1 496 918

7 5 2 52. . . .

. ; .pdo

− = × =

wheree2 is the edge distancep2 is the gauge = 45 + 45 + 6.9 = 96.9 mm

Hence, k1 = 2.5 and the design bearing resistance of one shear plane, Fb,Rd, is given by

F k f

dtb,Rd b u,c

c

M

kN . .

. . = = × × ×

×× =−

12

32 5 0 555 41016 10

1 2510 72 8α

γ

Bearing resistance of bolt group = 10Fb,Rd = 10 × 72.8 = 728 kN > VEd = 200 kN OK

SHEAR RESISTANCE OF CLEATS CONNECTED TO SUPPORTING COLUMN


Connections

For S275 steel

fu = 410 N mm−2 (Table 9.4)

fy = 275 N mm−2 (Table 9.4)

Shear resistance based on gross area of cleat, A v = tchc = 10 × 260 = 2600 mm2. Hence shear resistance of each legof cleat, Vpl,Rd, is

V

A fpl,Rd

v y

M

kN ( / )

( / )

. . = =

×× =−3 2600 275 3

1 0010 412 8

0

3

γ

Shear resistance based on net area of cleat, A v,net = tc(hc − nd0) = 10 × (260 − 5 × 18) = 1700 mm2. Shear resistanceof each leg of cleat, VRd,net, is

V

A fRd,net

v,net u,c

M

kN ( / )

( / )

. . = =

×× =−3 1700 410 3

1 2510 321 9

2

3

γ (critical)

Hence total shear resistance of both legs of cleats

= 2 × VRd,net = 2 × 321.9 = 643.8 kN > VEd = 200 kN OK

VEd

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432


Here αb is the smallest of


ed

ff

1

335

3 180 648

800410

1 95 1 0o

ub

u

. ; . ; .=×

= = =


SHEAR RESISTANCE OF BOLT GROUP CONNECTING CLEATS TO WEB OF SUPPORTED BEAM

Fv.Ed

Shear force per bolt in vertical direction, Fv,Ed, is

F

Vv,Ed

Ed kN = = =5

200

540

Maximum shear force on bolt assembly in horizontal direction, Fh,Ed, is

F

M

p

V s

ph,Ed

Ed Ed kN

= = =

××

=5 5

200 45

5 5036

1

The design shear force, Fv,Ed, is given by

Fv,Ed = [Fv,Ed2 + Fh,Ed

2]1/2 = [402 + 362]1/2 = 53.8 kN

Shear resistance per bolt, Fv,Rd, is 60.3 kN (see above).Since the bolts are in double shear the total shear resistance is

2Fv,Rd = 2 × 60.3 = 120.6 kN > FV,Ed = 53.8 kN OK

BEARING RESISTANCE OF WEB OF SUPPORTED BEAM

Fv.Ed

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433

Connections

Example 9.15 continuedHere k1 is the smallest of


2 8 1 7 2 86018

1 7 9 3 2 52. . . . . ; .edo

− = × − =

Hence, k1 = 2.5.

The design bearing resistance of one shear plane, Fb,Rd, is given by

F k f

dtb,Rd b u

w

M

. . ..

= = × × ××

× −1

2

32 5 0 648 41016 6 9

1 2510α

γ

= 58.6 kN > Fv,Ed = 53.8 kN OK

BLOCK TEARING

Ant = (a2 − 0.5do)tw = (35 − 0.5 × 18) × 6.9 = 179.4 mm2 (Fig. 9.12)

Anv = (L v + a1 − 4.5do)tw = (200 + 60 − 4.5 × 18) × 6.9 = 1235.1 mm2

V

f A f Aeff,1,Rd

u nt

M

y nv

M

( / )

..

( / ) .

.= + =

×+

γ γ2 0

3 430 179 41 25

275 3 1235 11 00

= 61713.6 + 196098.5 = 257812 N

= 257.8 kN > VEd = 200 kN OK

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Eurocode 6: Design of masonry structures

434

load-bearing members to prevent premature col-lapse of the structure and the measures required tolimit the spread of fire in masonry structures.

Part 2 provides guidance on, amongst otheraspects, the selection of mortars and masonry unitsfor various exposure conditions and applications.It is somewhat similar in scope to BS 5628: Part 3but the coverage in the European standard is lessextensive. This is intimated in clause 4 of theaccompanying National Annex which states that‘a standard comprising complementary and non-contradictory material taken from BS 5628–1, BS5628–2 and BS 5268–3 is in preparation’.

Part 3 contains simple rules for the design ofvarious types of unreinforced masonry walls includ-ing panel, shear and basement walls. The rulesare consistent with those provided in Part1.1 butwill lead to more conservative designs and wouldappear to have been included because they havetraditionally been used in some EC Member Statecountries. Neither Part 1.2 nor Part 3 of Eurocode6 is discussed in this chapter.

Part 1.1 of Eurocode 6, hereafter referred to asEC 6, was published as a preliminary standard,reference no: DD ENV 1996–1–1, in 1994 andthen in final form in 2004. It is expected to replaceBS 5628 by around 2010.

In common with the other structural Eurocodes,design of masonry structures cannot wholly beundertaken using EC 6. Reference will have to bemade to a number of other documents, notablyEN 1990 and Eurocode 1 for details of designphilosophy as well as rules for determining thedesign value of actions and combination of actionsas discussed below.

10.2 LayoutIn common with the other structural Eurocodes,EC 6 has been drafted by a panel of experts drawn

This chapter describes the contents of Part 1.1 ofEurocode 6, the new European standard for the designof masonry structures, which is expected to replace BS5628 by about 2010. The chapter highlights the principaldifferences between Eurocode 6: Part 1.1 and BS 5628and illustrates the new design procedures by means of anumber of worked examples on single leaf and cavitywalls, with and without stiffening piers, subject to eithervertical or lateral loading or a combination of both.

10.1 IntroductionEurocode 6 is the new European standard forthe design of masonry structures. It applies to thedesign of buildings and civil engineering works in,predominantly, unreinforced and reinforced ma-sonry. It is based on limit state principles and ispublished in four parts as shown in Table 10.1.

Part 1.1 of Eurocode 6 which is the focus ofthe discussion in this chapter provides the basicinformation necessary for the design of masonrystructures. It deals with the material properties andgives detailed rules which are mainly applicableto ordinary buildings. The corresponding BritishStandard is BS 5628: Parts 1 and 2.

Part 1.2 deals with the accidental event of fire andaddresses, amongst other aspects, fire protection of

Table 10.1 Scope of Eurocode 6: Design ofMasonry Structures

Part Subject

1.1 General rules for reinforced and unreinforcedmasonry

1.2 Structural fire design2 Design consideration, selection of materials and

execution of masonry3 Simplified calculation methods for unreinforced

masonry structures

Chapter 10

Eurocode 6:Design of masonrystructures

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435

equivalent and do not adversely affect other designrequirements. It is worth noting, however, that ifan alternative application rule is used the resultingdesign will not be deemed Eurocode compliant.


Like the other structural Eurocodes, Eurocode 6allows some parameters and design methods to bedetermined at the national level. Where a nationalchoice is allowed this is indicated in a note in thenormative text under the relevant clause. The notemay include the recommended value of the par-ameter, or preferred method, etc., but the actualvalue, methodology, etc., to be used in a particularMember State country is given in the appropri-ate National Annex. The recommended values ofthese parameters and design method/procedures arecollectively referred to as Nationally DeterminedParameters (NDPs). The NDPs determine variousaspects of design but perhaps most importantlythe level of safety of structures during execution(i.e. construction/fabrication) and in-service, whichremains the responsibility of individual nations.

The UK National Annexes for all four parts ofEurocode 6 were published in 2005 and the dis-cussion and the worked examples in this chapterare based on the material in EC 6 and the NDPsin the accompanying National Annex.

10.5 SymbolsChapter 1 of EC 6 lists the symbols used in thedocument. Those relevant to this discussion arereproduced below.

GEOMETRIC PROPERTIES:t thickness of a wall or leafh height of panel between restraintsl length of wall between restraintsA gross cross-sectional area of a wallZ elastic sectional modulustef effective thickness of a wallhef effective height of a wallρt stiffness coefficientehe eccentricity at the top and bottom of a

wall, resulting from horizontal loadsehm eccentricity at the middle of a wall,

resulting from horizontal loadsei eccentricity at the top and bottom of a walleinit initial eccentricity

Symbols

from the various EC Member State countries. Thebase documents for the first draft of EC 6 wereReport 58: International recommendations for thedesign of masonry structures and Report 94: Interna-tional report for design and erection of unreinforced andreinforced masonry structures, published in 1980 and1987 respectively, prepared by CIB (Council forInternational Building) committee W23. The fol-lowing subjects are covered in EC 6:

Chapter 1: GeneralChapter 2: Basis of designChapter 3: MaterialsChapter 4: DurabilityChapter 5: Structural analysisChapter 6: Ultimate limit statesChapter 7: Serviceability limit statesChapter 8: DetailingChapter 9: Execution

In addition, supplementary information is pro-vided in a number of annexes. All the annexes inEC 6 are labelled ‘informative’. As explained insection 7.5.4 of this book, this signifies that thismaterial does not have any status and is includedmerely for information.

As can be appreciated from the above contents list,the organisation of material is different to that usedin BS 5628 but follows the layout adopted in theother structural Eurocodes. Generally, the designrules in EC 6 are sequenced on the basis of actioneffects rather than type of member, as in BS 5628.

This chapter briefly describes the contents ofEC 6, insofar as it is relevant to the design of singleleaf and cavity walls, with or without stiffening piers,subject to vertical and/or lateral loading.

10.3 Principles/Application rules(Cl. 1.4, EC 6)

As with the other structural Eurocodes and for thereasons discussed in Chapter 7, the clauses in EC5 have been divided into Principles and Applica-tion rules. Principles comprise general statements,definitions, requirements and models for which noalternative is permitted. Principles are indicatedby the letter P after the clause number. The Appli-cation rules are generally recognised rules whichfollow the statements and satisfy the requirementsgiven in the Principles. The absence of the letter Pafter the clause number indicates an applicationrule. The use of alternative application rules to thoserecommended in the Eurocode is permitted pro-vided it can be shown that the alternatives are at least

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436

ek eccentricity due to creepem eccentricity due to loadsemk eccentricity at the middle of the wallSR slenderness ratioΦ capacity reduction factorΦi capacity reduction factor at the top or

bottom of the wallΦm capacity reduction factor at mid-height of

the wall

ACTIONS AND PARTIAL SAFETY FACTORS:Gk characteristic permanent actionQk characteristic imposed actionWk characteristic wind loadγ f, γF partial factor for actionsγM partial factor for materials

COMPRESSION:E short-term secant modulus of elasticity of

masonryfb normalised mean compressive strength of

a masonry unitfd design compressive strength of masonryfk characteristic compressive strength of

masonryfm compressive strength of masonry mortarfk characteristic compressive strength of

masonryN design vertical actionNRd design vertical resistance of a masonry wall

FLEXURE:fxk1 characteristic flexural strength of masonry

with plane of failure parallel to bed jointsfxk2 characteristic flexural strength of masonry

with plane of failure perpendicular to bedjoints

fxd design flexural strength of masonryα1,2 bending moment coefficientsµ orthogonal ratio of the flexural strengths

of masonryMEd design value of the applied momentMRd design value of moment of resistanceMEd1 design applied moment with plane of

failure parallel to bed jointMEd2 design applied moment with plane of

failure perpendicular to bed jointMRd1 design moment of resistance with plane of

failure parallel to bed jointMRd2 design moment of resistance with plane of

failure perpendicular to bed joint

10.6 Basis of designLike BS 5628, EC 6 is a limit state code. Thetwo principal categories of limit states relevantto the design of masonry structures are durabilityand strength. Design for durability is discussed inchapter 4 of EC 6 and largely relates to the selec-tion of masonry units and mortars for particularstructure types and exposure classes. The codeprovisions relevant to this aspect of design are dis-cussed in section 10.8. The design rules dealingwith ultimate limit states are given in chapter 6 ofEC 6. Only those rules relevant to the design ofunreinforced masonry walls subjected to eithermainly vertical or lateral loading are discussed inthis chapter. Generally, in order to assess the effectof these loading conditions on masonry structuresthe designer will need to estimate

(a) the design values of actions(b) the design strength of materials.

10.7 ActionsAction is the Eurocode terminology for loads andimposed deformations. Permanent actions, G, areall the fixed loads acting on the structure, includingfinishes, immovable partitions and the self-weightof the structure. Variable actions, Q, include theimposed, wind and snow loads. Clause 2.3.1(1) ofEC 6 recommends that the values of characteristicpermanent, Gk, and variable, Qk, actions shouldbe obtained from the relevant parts of Eurocode 1:Actions on structures. Guidance on determining thedesign value of actions and combination of actionsis given in EN 1990: Basis of structural design. Thesedocuments and topics are briefly discussed insection 8.5 of this book. As noted there, the designvalue of an action (Fd) is obtained by multiplyingthe representative value (Frep) by the appropriatepartial safety factor for the action (γf):

Fd = γfFrep (10.1)

Table 10.2 shows the relevant partial safety fac-tors for the ultimate limit state of strength. Othersafety factors will apply in other design situations.For example the partial factors for the ultimatelimit states of equilibrium are shown in Table 8.6.In equation 10.1, Frep is generally taken as the char-acteristic value of a permanent or variable action(i.e. Fk). Assuming that the member being designedis subjected to one (or more) permanent actionsand one variable action, i.e. load combination 1 inTable 10.2, the partial safety factor for permanent

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437

actions, γG can conservatively be taken as 1.35 andfor the variable action, γQ as 1.5. As discussed insection 8.5.3, it is possible to improve structuralefficiency by using expressions 6.10a and 6.10bof EN 1990 (respectively load combination 3(a)and 3(b)/3(c) in Table 10.2) to estimate the designvalues of actions but the value of 1.35 for γG isconservative and used throughout this chapter.

10.8 Design compressive strengthThe design compressive strength of masonry, fd, isgiven by

f

fd

k

M

=γ

(10.2)

wherefk is the characteristic compressive strength of

masonryγM is the partial factor for materials.

The characteristic compressive strength ofmasonry is a function of the following product/material characteristics:

• Group number of masonry unit• Normalised mean compressive strength of

masonry unit• Compressive strength of the mortar.

The partial factor for materials is a function of thefollowing aspects:

• Category of (masonry unit) manufacturingcontrol

• Class of execution control.






(a) 1.35 1.0 1.5ψ0,1 0 1.5ψ0,2

(b) 1.35/1.35ξ 1.0 1.5 0 1.5ψ0

(c) 1.35/1.35ξ 1.0 1.5ψ0 0 1.5

Some of this information is provided by themanufacturer of the masonry units whereas theothers are determined using the guidance given inEurocode 6. Before describing how the character-istic strength of masonry and the partial factor formaterials are actually evaluated, the following brieflyreviews the purpose of each of the above parameters.

10.8.1 GROUP NUMBEROne of the stated aims of the Single Europe Act isthat there should be no barriers to trade, which hasgenerally been interpreted as meaning that designrules must not disadvantage available products fromEC countries. Yet there is a wide variety of ma-sonry units used throughout Europe; see Fig. 10.1for examples. The units differ in many respectsincluding percentage, size and orientation of voidsor perforations, and the thickness of webs and shells,etc. Thus, the EC 6 drafting panel had to developmethods that would allow the majority of masonryproducts currently available throughout Europe tobe used in design.

The method actually developed involved produc-ing European specifications for common types ofmasonry unit: clay, calcium silicate, aggregate con-crete, autoclaved aerated concrete, manufacturedstone and natural stone. These specifications use adeclaration system to specify product characteristics,each of which is supported with an appropriate testmethod. In turn, the product characteristics areused to assign masonry units with similar charac-teristics into one of four groups: 1, 2, 3 and 4, withthe superior structural use masonry units beingplaced in Group 1. In general, Group 1 units haveno more than 25 per cent voids. Group 2 clay and

Design compressive strength

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438

A number of procedures for conditioning ofmasonry units prior to testing are outlined in EN772-1: Methods of test for masonry units: Determinationof compressive strength. This standard advises thatthe conditioning factor for air-dried units is 1.0.Masonry units manufactured to British Standardsare wet strengths in which case the recommendedvalue of the conditioning factor is 1.2. Values forthe shape factor, δ, are given in Table A1 of EN772–1, to allow for the height and width of units.For example, for 102.5 mm × 65 mm bricks δ =0.85, and for 215 mm (height) × 100 mm (width/thickness) blocks, δ = 1.38.

10.8.3 COMPRESSIVE STRENGTH OF MORTARTable 10.4 shows the masonry mortar mixes recom-mended for use in the UK to achieve the appro-priate strength given in EC 6. As will be noted thechoice and designation of masonry mortars in EC6 and BS 5628 are identical and therefore for agiven application or exposure similar mortars maybe specified. According to clause 3.2.2 of EC 6,masonry mortars may be specified by compressivestrength, expressed as the letter M followed by thecompressive strength in Nmm−2, e.g. M4, or mixproportion, e.g. 1:1:6 signifies the cement-lime-sandproportions by volume. The latter has the advant-age, however, that it will produce mortars of knowndurability and should generally be used in practice.

calcium silicate units have between 25–55 per centvoids and aggregate concrete units between 25–60per cent voids. Table 10.3 gives further detailsof the requirements for unit groupings. All UKbricks currently manufactured to British Standards,including frogged and perforated bricks, fit intoGroup 1 unit specification although the authorunderstands some Group 2 units are becomingavailable. Cellular and hollow blocks fit intoGroup 1 or Group 2 unit specification, dependingon the void content. Masonry units which fall withinGroups 3 and 4 have not historically been used inthe UK. The manufacturer will normally declarethe group number appropriate to his unit.

10.8.2 NORMALISED COMPRESSIVE STRENGTHA difference in the size of units available through-out Europe and differences in test procedures hasmeant that the compressive strength of masonryunits had to be normalised.

The normalised compressive strength, fb, is thecompressive strength converted to the air driedcompressive strength of an equivalent 100 mm wide× 100 mm high unit of the same material. Thenormalised compressive strengths of masonry, fb,is given by

fb = conditioning factor × shape factor ×declared mean compressive strength (10.3)

Solid Unit

Vertically Perforated Unit Vertically Perforated Unit

Frogged Unit Vertically perforated Unit

Fig. 10.1 Examples of high density (HD) clay masonry units (Fig. 3, EN 771-1).

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439

Table 10.3 Geometrical requirements for groupings of clay bricks and concrete blocks (based onTable 3.1, EC 6)

Materials and limits for masonry units

Group 1 Group 2 Group 3 Group 4(All materials)

Units Vertical holes Vertical holes

Volume of all holes(% of the gross volume)

Volume of any hole(% of the gross volume)

Declared values ofthickness of weband shell (mm)

Declared value ofcombined thicknessof webs and shells(% of the overall width)

≤ 25

≤ 12.5

No requirement

No requirement

ClayConcrete

Clay

Concrete

ClayConcrete

ClayConcrete

> 25; ≤ 60> 25; ≤ 60

Each of multipleholes ≤ 2gripholes up toa total of 12.5

Each of multipleholes ≤ 30gripholes up toa total of 30

Web Shell≥ 5 ≥ 8≥ 15 ≥ 18

≥ 16≥ 18

≥ 25; ≤ 70> 25; ≤ 70

Each of multipleholes ≤ 2gripholes up toa total of 12.5

Each of multipleholes ≤ 30gripholes up toa total of 30

Web Shell≥ 3 ≥ 6≥ 15 ≥ 15

≥ 12≥ 15

> 25; ≤ 70> 25; ≤ 50

Each of multipleholes ≤ 30

Each of multipleholes ≤ 25

Web Shell≥ 5 ≥ 6≥ 20 ≥ 20

≥ 12≥ 45

Table 10.4 Types of mortars (Table 2 of National Annex to EC 6)

Compressive Prescribed mortars (proportion of materials by volume) Mortarstrength class designation

Cement-lime-sand Cement-sand with Masonry Masonrywith or without or without air cement1-sand cement2-sandair entrainment entrainment

M12 1:0 to 1/4:3 – – – (i)M6 1:1/2:4 to 41/2 1:3 to 4 1:21/2 to 31/2 1:3 (ii)M4 1:1:5 to 6 1:5 to 6 1:4 to 5 1:31/2 to 4 (iii)M2 1:2:8 to 9 1:7 to 8 1:51/2 to 61/2 1:41/2 (iv)

Notes:1Masonry cement with organic filler other than lime2Masonry cement with lime

Design compressive strength

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10.8.4 CATEGORY OF UNIT MANUFACTURINGCONTROL

Units manufactured in accordance with Europeanspecifications can be further classified as belongingto Category I or Category II depending on themanufacturing control. Category I units are thosewhere the manufacturer operates a quality-controlscheme and the probability of the units not reach-ing the declared compressive strength is less than 5per cent. Masonry units not intended to complywith the Category I level of confidence are classi-fied as Category II. BS 5628 has adopted the samesystem of classifying masonry units (Table 5.10).It is the manufacturer’s responsibility to declarethe category of the masonry unit supplied.

10.8.5 CLASS OF EXECUTION CONTROLEC 6 allows for up to five classes of executioncontrol but, as in BS 5628, only two classes areused in the UK National Annex, namely, 1 and 2.According to Table 1 of the National Annex,Class 1 execution control should be assumedwhenever the work is carried out following therecommendations for workmanship in EN 1996:Part 2 (EC 6–2), including appropriate supervisionand inspection, and in addition:

(a) the specification, supervision and control ensurethat the construction is compatible with theuse of the appropriate safety factors given inEC 6;

(b) the mortar conforms to BS EN 998–2, if itis factory made mortar, or if it is site mixedmortar, preliminary compressive strength testscarried out on the mortar to be used, inaccordance with BS EN 1015–2 and BS EN

1015–11, indicate conformity to the strengthrequirements given in EC 6 and regular testingof the mortar used on site, in accordance withBS EN 1015–2 and BS EN 1015–11, showsthat the strength requirements of EC 6 arebeing maintained.

Class 2 execution control should be assumedwhenever the work is carried out following therecommendations for workmanship in EC 6–2,including appropriate supervision.

Class 1 execution control corresponds to the‘special’ category of construction control used inBS 5628 and Class 2 to the ‘normal’ category.

10.8.6 CHARACTERISTIC COMPRESSIVESTRENGTH OF MASONRY

The characteristic compressive strength ofunreinforced masonry, fk, built with general-purposemortar can be determined using the followingexpression

fk = K f b0.7f m

0.3 (10.4)

wherefm is the compressive strength of general-

purpose mortar but not exceeding 20 Nmm−2

or 2 fb, whichever is the smallerfb is the normalised compressive strength of the

masonry unitsK is a constant obtained from Table 10.5.

The compressive strength of general-purposemortar is obtained from Table 2 of the NationalAnnex to EC 6, reproduced here as Table 10.4.This assumes that mortar designations (i), (ii),(iii) and (iv) (as defined in Table 10.4) have

Table 10.5 Values of K (based on Table 4 of the National Annexto EC 6)

Masonry unit Values of K for generalpurpose mortar

Clay Group 1 0.50Group 2 0.40

Calcium silicate Group 1 0.50Group 2 0.40

Aggregate concrete Group 1 0.55Group 1a (units laid flat) 0.50Group 2 0.52

aif units contain vertical voids multiply K by (100-n)/100, where n is the percentageof voids, maximum 25%

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441

10.10 Design of unreinforcedmasonry walls subjectedto vertical loading

Having discussed the basics, the following outlinesEC 6 rules for the design of vertically loaded wallsas set out in section 6.1. The approach is verysimilar to that in BS 5628 and principally involveschecking that the design value of the vertical load,NEd, is less than or equal to the design value of thevertical resistance of the wall, NRd, i.e.

NEd ≤ NRd (10.5)

According to clause 6.1.2.1 of EC 6, the designvertical load resistance of a single leaf unreinforcedmasonry wall per unit length, NRd, is given by

N

tfRd

i,m k

M

=Φ

γ(10.6)

whereΦi,m is the capacity reduction factor, Φi or Φm,

as appropriatet is the thickness of the wallfk is the characteristic compressive strength of

masonry obtained from equation 10.4γM is the material factor of safety for masonry

determined from Table 10.6.

The factor Φi which relates to the top or bottomof the wall is given by

Φi = 1 − 2e i /t (10.7)

where e i is the eccentricity at the top or bottom ofthe wall given by

e i = Mi /Ni + ehe ± e init ≥ 0.05t (10.8)

whereM i is the design bending moment at the top or

bottom of the wall

compressive strengths of, respectively, 12 Nmm−2,6 Nmm−2, 4 Nmm−2 and 2 Nmm−2. Unlike BS 5628,EC 6 does not differentiate between site and lab-oratory strengths.

10.8.7 PARTIAL FACTOR FOR MATERIALS, γMTable 10.6 shows the values of the partial factorsfor material properties for the ultimate limit stategiven in Table 1 of the National Annex to EC 6.As can be seen they are primarily a function of theCategory of unit, Class of execution control andthe state of stress. Comparison with the correspond-ing values in BS 5628 (Table 5.10) shows that thevalues in EC 6 are somewhat lower.

10.9 DurabilityAs previously noted the other limit state which mustbe considered in masonry design is durability. Thus,masonry units and mortars should be sufficientlydurable to resist the relevant exposure conditionsfor the intended life of the structure, normallyassumed to be 50 years for buildings. Table 10.7shows the five main exposure classes and sub-classesrelevant to masonry design mentioned in EC 6.It also gives examples of structures that mayexperience these conditions. Tables B1 and B2 ofEC 6–2 gives details of acceptable specifications ofmasonry units and mortars appropriate to theseexposures classes. Clause 3 of the associatedNational Annex recommends that this informationis not used, however. As previously noted, an NCCI(Non-Contradictory Complementary Information)on the selection of materials is in preparationand in the interim the guidance in Table 12 ofBS 5628 (Table 5.7), which is both extensive andrelevant to UK exposure conditions and practice,should be used.

Design of unreinforced masonry walls subjected to vertical loading

Table 10.6 Values of γM for ultimate limit state (based onTable 2.3 of EC 6 and Table 1 of the National Annex to EC 6)

Class of execution control

1 2

When in a state of direct or flexural compressionUnreinforced masonry made with:

units of category I 2.3 2.7units of category II 2.6 3.0

When in a state of flexural tensionunits of category I and II 2.3 2.7

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Table 10.7 Classification of conditions of exposure of masonry (Table A1, EC 6)

Class

MX1

MX2MX2.1

MX2.2

MX3MX3.1

MX3.2

MX4

MX5

Typical examples

Interior walls of building for normal habitationand for offices; rendered exterior walls largelyunexposed to rain and isolated from damp

Laundry room interior walls; exterior wallssheltered by overhanging eaves; masonry belowfrost zone in well-drained non-aggressive soil.

Exterior walls with cappings of flush eaves;parapets; freestanding walls in the ground orunderwater.

As class MX2.1 but susceptible to freeze/thaw

As class MX2.1 but susceptible to freeze/thaw

Masonry in coastal regions or adjacent to roadsthat are salted

Masonry in contact with natural soils or madeground or groundwater, where moisture andsulphates are present; masonry in contact withacidic soil or contaminated ground conditions;masonry in industrial areas where airborneaggressive chemicals are present.

Description

Dry environment

Exposed to moisture or wettingExposed to moisture but not freeze/thaw cyclingor external sources of sulphates or aggressivechemicals

Exposed to severe wetting but not freeze/thawcycling or external sources of sulphates oraggressive chemicals

Exposed to wetting plus freeze/thaw cyclesExposed to moisture or wetting and freeze/thawcycling but not sulphates or aggressive chemicals

Exposed to severe wetting and freeze/thaw cyclingbut not sulphates or aggressive chemicals

Exposed to saturated salt air, sea water ordeicing salts

In an aggressive chemical environment

Ni is the design vertical load at the top orbottom of the wall

ehe is the eccentricity at the top or bottom ofthe wall, if any, resulting from horizontalloads e.g. wind

einit is the accidental eccentricity resulting fromconstruction inaccuracies and can be takenas hef/450 (10.9)

in which hef is the effective height of the wall (see10.9.1).

The value of the capacity reduction factor Φm,which relates to the mid-height of the wall, isestimated using either Fig. G.1 of EC 6 or thefollowing expressions from Annex G of EC 6. Theseequations are valid for masonry of modulus ofelasticity, E = 1000fk although the general form ofthe equations are applicable to masonry of anymodulus:

Φm =−

A eu

12

2

(10.10)

where

A

et1 1 2 = − mk (10.11)

u

ht

et

=

−

−

ef

ef

mk

2

23 37(10.12)

in whichhef is the effective height of the walltef is the effective thickness of the wallemk is the eccentricity in the middle of the wall

= em + ek (10.13)

whereek is the eccentricity due to creep and is equal

to zero where SR < 27 (Cl. 2.14 of NationalAnnex)

em = Mm/Nm + ehm ± einit ≥ 0.05t (10.14)

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in whichMm is the design bending moment in the

middle of the wallNm is the design vertical load in the middle of

the wallehm is the greatest eccentricity in the middle

one-fifth height of the wall, if any, due tolateral loads

einit is the initial eccentricity.

A simplified sub-frame analysis proceduredescribed in Annex C of EC 6 can be used tocalculate the values of the design bending momentsand thence the corresponding values of ei and emk.The smaller of Φi and Φm is used to estimate thevertical design resistance of the wall.

Clause 4.4.2(3) notes that where the cross-sectional area of a wall is less than 0.1 m2, thecharacteristic compressive strength of masonryshould be multiplied by (0.7 + 3A), where A is theloaded horizontal gross cross-sectional area of themember, expressed in square metres. Unlike BS5628, there is no enhancement factor for narrowwalls used in EC 6; rather, there is a 0.8 reductionfactor for walls with collar joints in their middle.

10.10.1 EFFECTIVE HEIGHT, hef (CL. 5.5.1.2, EC 6)The effective wall height is a function of the actualwall height, h, and end/edge restraints. It can betaken as

hef = ρnh (10.15)

ρn is a reduction factor where n = 2, 3 or 4 depend-ing on the number of restrained and stiffened edges.Thus, n = 2 for walls restrained at the top andbottom only, n = 3 for walls restrained top andbottom and stiffened on one vertical edge withthe other vertical edge free and n = 4 for wallsrestrained top and bottom and stiffened on twovertical edges.

For walls with simple resistance (Fig. 5.10) atthe top and bottom, ρ2 = 1 as in BS 5628. Forwalls with enhanced resistance (Fig. 5.11) at thetop and bottom ρ2 = 0.75 as in BS 5628. ρn

assumes other values if the wall is additionallystiffened along one or more vertical edges. Seeclause 4.4.4.3 of EC 6 for details.

10.10.2 EFFECTIVE THICKNESS, tef (CL. 5.5.1.3,EC 6)

The effective thickness of a single leaf wall is equalto the actual thickness, but for cavity walls in whichthe leaves are connected by suitable wall ties it isgenerally given by

t k t tef tef = +13

233 (10.16)

wheret1 and t2 are the thicknesses of the two leaves.ktef is the ratio of E values of the two

leaves of the wall = E1/E2 = 1 (Cl 2.13,National Annex to EC 6)

The effective thickness of a wall stiffened by piersis given by

tef = ρtt (10.17)

wheret is the actual thickness of the wallρt is a coefficient obtained from Table 5.1

(reproduced here as Table 10.8) incombination with Fig. 5.2 (reproducedhere as Fig. 10.2), both of EC 6.

Pierthickness

Pierwidth

Thicknessof wall

Pier spacing

Fig. 10.2 Diagrammatic view of the definitions used inTable 10.6 (Fig. 5.2, EC 6).

Table 10.8 Stiffness coefficient, ρt, for wallsstiffened by piers, see Fig. 10.2 (Table 5.1,EC 6)

Ratio of pier Ratio of pier thickness tospacing (centre actual thickness of wall toto centre) to which it is bondedpier width

1 2 3

6 1.0 1.4 2.010 1.0 1.2 1.420 1.0 1.0 1.0

Note: Linear interpolation between the values in the table ispermitted


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Example 10.1 Design of a loadbearing brick wall (EC 6)The internal load-bearing brick wall shown in Fig. 10.3 supports an ultimate axial load of 140 kN per metre runincluding self-weight of the wall. The wall is 102.5 mm thick and 4 m long. Assuming that the manufacturing controlof the units is category II and the execution control is class 2, design the wall.

LOADINGUltimate design load, NEd = 140 kN m−1 = 140 N mm−1

SLENDERNESS RATIO (SR)Concrete slab provides ‘enhanced’ resistance to wall, therefore effective height of the wall, hef, is given by

hef = ρnh = 0.75 × 2800 = 2100mm

Effective thickness of wall, tef = actual thickness (single leaf) = 102.5 mm

Slenderness ratio, SR =

ht

ef

ef

.

=2100102 5

= 20.5 < 27 ⇒ the effects of creep can be ignored, i.e. ek = 0

CAPACITY REDUCTION FACTOREccentricity at the top and bottom of wallMid/Nid = 0 since the wall is axially loadedehe = 0, since there are no horizontal loads presenteinit = hef/450 = 2100/450 = 4.7mm

The eccentricity of the design vertical load at the top and bottom of the wall, ei, is given by

ei = M i/Ni + ehi ± einit = 0 + 0 + 4.7 ≥ 0.05t = 0.05 × 102.5 = 5.125 mm

Hence ei = 5.125 mm = 0.05tThe corresponding capacity reduction factor, Φi = 1 − 2ei/t = 1 − 2 × (0.05t)/t = 0.9

Eccentricity in middle of the wallMmd/Nmd = 0

ehm = 0, since there are no horizontal loads present

einit = hef/450 = 2100/450 = 4.7 mm

Eccentricity of the design vertical load in the middle of the wall, em, is given by

em = Mmd/Nmd + ehm ± einit = 0 + 0 + 4.7 ≥ 0.05t = 0.05 × 102.5 = 5.125 mm

Total eccentricity in the middle of the wall, emk = em + ek = 5.125 + 0 = 0.05tModulus of elasticity of masonry, E = 1000fk (Cl. 2.9 of National Annex to EC 6)

u

ht

et

tt

.

( . )

.=−

−=

−

−=

ef

ef

mk

2

23 37

20 5 2

23 370 05

0 875

Fig. 10.3

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445

A

et

tt1 1 2 1 2

0 050 9

. .= − = − × =mk

Φm (critical) . . .

= = × =− −A e e

u

12

0 875

2

2 2

0 9 0 61

DESIGN VERTICAL LOAD RESISTANCE OF WALL (NRd)

Modification factorSmall plan area modification factor does not apply since horizontal cross-sectional area of wall, A = 0.1025 × 4.0 =0.41 m2 > 0.1 m2.

Safety factor for materials (γγγγγM)Since manufacturing control is Category II and execution control is Class 2, from Table 10.6 γM = 3.0

Design vertical load resistance

N

tf fRd

i,m k

M

k . .

.= =

×Φγ

0 61 102 53 0

DETERMINATION OF fk


NRd ≥ NEd

0.61 102.5 3.0

k× ×≥

f 140

Hence

fk Nmm

. . ≥ =

140

20 86 8 2

SELECTION OF BRICK AND MORTAR TYPEAssuming the wall is made of Group 1 clay bricks of standard format size laid in M4 general purpose mortar

⇒ K = 0.5 (Table 10.5)

⇒ fm = 4 Nmm−2 (Table 10.4)

⇒ shape factor = 0.85 (section 10.7.2)

⇒ conditioning factor = 1 (assuming units are manufactured to EN 771-1)

The characteristic strength of masonry, fk, is given by

f k = Kf b0.7f m

0.3

6.8 = 0.5 × f b0.7 × 40.3

Therefore the normalised compressive strength of the units, fb ≥ 23 N mm−2

The declared mean compressive strength of the units is given by

fb = conditioning factor × shape factor × declared mean compressive strength23 = 1.0 × 0.85 × declared mean compressive strength

Therefore the required declared air dried mean compressive strength of units ≥ 27 N mm−2

The actual brick type that will be specified on the working drawings will depend not only upon the structuralrequirements but also on aesthetics, durability, buildability and cost, amongst other considerations. In this particularcase the designer may specify the minimum requirements as declared air dried mean brick strength: ≥ 27 N mm−2,mortar mix: 1:1:6, frost resistance/soluble salt content: F0/S1 (see Tables 5.1 and 5.2). Assuming that the wall will beplastered on both sides, appearance of the brick will not need to be specified.



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Example 10.2 Design of a brick wall with ‘small’ plan area (EC 6)Redesign the wall in Example 10.1 assuming that it is only 0.9 m long.

The calculations for this case are essentially the same as for the 4 m long wall except for the fact the plan areaof the wall, A, is now less than 0.1m2, being equal to 0.1025 × 0.9 = 0.09225 m2. The ‘plan area’ modification factoris equal to

(0.70 + 1.5A) = 0.7 + 1.5 × 0.09225 = 0.83

Therefore the modified characteristic compressive strength of masonry is given by

0.83fk

From the above γM = 3.0 and Φm = 0.61. Hence, the required characteristic compressive strength of masonry, fk, isgiven by

0 61 102 5 0 83140

. . .

× ×≥

fk

3.0

Hence

fk

. ≥ =

140

17 38.1 Nmm−2

Again assuming the wall is made of Group 1 clay bricks of standard format size laid in M4 general purpose mortar

⇒ K = 0.5

⇒ fm = 4 Nmm−2

⇒ shape factor = 0.85

⇒ conditioning factor = 1.0

f k = Kf b0.7f m

0.3

8.1 = 0.5 × f b0.7 × 40.3

⇒ fb ≥ 29.5 Nmm−2

fb = conditioning factor × shape factor × declared mean compressive strength29.5 = 1.0 × 0.85 × declared mean compressive strength

Therefore the declared air dried mean compressive strength of units ≥ 35 Nmm−2

Hence it can be immediately seen that walls having similar construction details but a plan area of < 0.1 m2 will havea lower load-carrying capacity.

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ASSUMING ‘ENHANCED’ RESISTANCE

Characteristic compressive strength (fk)Conditioning factor = 1 since the declared strengths are on air dried units.Shape factor = 0.85 since the bricks are standard format size.Declared mean compressive strength of bricks is 30 Nmm−2.Normalised compressive strength, fb, is given by

fb = conditioning factor × shape factor × declared mean compressive strength= 1.0 × 0.85 × 30 = 25.5 Nmm−2

From Table 10.4, 1:1:6 mix corresponds to a grade M4 mortar. The characteristic compressive strength of masonry, fk,is given by

fk = Kf b0.7f m

0.3 = 0.5 × 25.50.7 × 40.3 = 7.3 Nmm−2

Capacity reduction factor (S)

Slenderness ratio (SR)With ‘enhanced’ resistance the effective height of wall, hef, is given by

hef = ρnh = 0.75 × 3500 = 2625 mm

Pier spacingPier width

.= =4500440

10 2

Pier thicknessThickness of wall

.= =440215

2 0

Hence from Table 10.8, stiffness coefficient ρt = 1.2. The effective thickness of the wall, tef, is equal to

tef = ρtt = 1.2 × 215 = 258 mm

SR ef

ef

= =ht

2625258

= 10.2 < 27 ⇒ the effects of creep may be ignored, i.e. ek = 0

Example 10.3 Analysis of brick walls stiffened with piers (EC 6)A 3.5 m high wall shown in cross-section Fig. 10.4 is constructed from clay bricks of standard format size having adeclared air dried mean compressive strength of 30 Nmm−2 laid in a 1:1:6 mortar. Calculate the ultimate load bearingcapacity of the wall assuming the partial safety factor for materials is 3.0 and the resistance to lateral loading is (A)‘enhanced’ and (B) ‘simple’.

Fig. 10.4

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Capacity reduction factorEccentricity at the top and bottom of the wall, ei

Mi/N = 0 since the wall is axially loaded

ehi = 0, since there are no horizontal loads present

einit = hef/450 = 2625/450 = 5.83 mm

ei = M i /Ni + ehi ± einit ≥ 0.05t = 0.05 × 215 = 10.75 mm

= 0 + 0 + 5.83

Hence ei = 0.05t and Φi = 1 − 2e i/t = 1 − 2 × (0.05t)/t = 0.9

Eccentricity of the design load in the middle of the wall, emk

Mmd/Nmd = 0


e init = hef/450 = 2625/450 = 5.83 mm

em = Mmd/Nmd + ehm ± einit ≥ 0.05t = 0.05 × 215 = 10.75 mm

= 0 + 0 + 5.83

emk = em + e k = 0.05t + 0 = 0.05t

Hence emk = 0.05t

Modulus of elasticity of masonry, E = 1000fk (Cl. 2.9 of National Annex to EC 6)

u

ht

et

tt

.

( . )

.=−

−=

−

−=

ef

ef

mk

2

23 37

10 2 2

23 370 05

0 388

A

et

tt1 1 2 1 2

0 050 9

. .= − = − × =mk


= = =− −A e e

u

12

0 388

2

2 2

0 9 0 83

Design resistance of wall (NRd)

Safety factor for material (sM)

γM = 3.0

Design resistance of wall

N

tfRd

i,m k

M

Nmm . .

. = =

× ×= −Φ

γ0 83 215 7 3

3 0434 1 run of wall = 434 kNm−1 run of wall

Hence the ultimate load capacity of wall, assuming enhanced resistance, is 434 kNm−1 run of wall.

ASSUMING ‘SIMPLE’ RESISTANCE

Characteristic compressive strength (fk)As before, fk = Kf b

0.7f m0.3 = 0.5 × 25.50.7 × 40.3 = 7.3 Nmm−2


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Capacity reduction factor (S)Slenderness ratio (SR)Since the end restraints are ‘simple’

hef = actual height = 3500 mmtef = 258 mm (as above)

Hence

SR ef

ef

.= = =ht

3500258

13 6 < 27 ⇒ the effects of creep may be ignored, i.e. ek = 0

Capacity reduction factor

Eccentricity at the top and bottom of the wall, ei

Mi/N = 0 since the wall is axially loaded


einit = hef/450 = 3500/450 = 7.77 mm

ei = M i/Ni + ehi ± e init ≥ 0.05t = 0.05 × 215 = 10.75 mm

= 0 + 0 + 7.7 = 7.7 mm

Hence e i = 0.05t and Φi = 1 − 2e i/t = 1 − 2 × (0.05t)/t = 0.9


Mmd/Nmd = 0


einit = hef/450 = 3500/450 = 7.77 mm

em = Mmd/Nmd + ehm ± einit ≥ 0.05t = 0.05 × 215 = 10.75 mm

= 0 + 0 + 5.83 = 5.83 mm

emk = em + e k = 0.05t + 0 = 0.05tHence emk = 0.05t

Assuming E = 1000fk (as above), implies that

u

ht

et

tt

.

( . )

.=−

−=

−

−=

ef

ef

mk

2

23 37

13 6 2

23 370 05

0 548

A

et

tt1 1 2 1 2

0 050 9

. .= − = − × =mk


= = =− −A e e

u

12

0 548

2

2 2

0 9 0 77

Design resistance of wall (NRd)The design vertical load resistance of the wall, NRd, is given by:

N

tfRd

i,m k

M

Nmm . .

. = =

× ×= −Φ

γ0 77 215 7 3

3 0402 1 run of wall = 402 kNm−1 run of wall

Hence it can be immediately seen that, all other factors being equal, walls having simple resistance have a lowerresistance to failure.



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Example 10.4 Design of a cavity wall (EC 6)A cavity wall of length 6m supports the loads shown in Fig. 10.5. The inner leaf is built using solid concrete block oflength 440 mm, width 100 mm and height 215 mm, faced with plaster, and the outer leaf from standard format claybricks. Design the wall assuming that the execution control is Class 1 and that the manufacturing control of theconcrete blocks is Category II. The self-weight of the blocks and plaster can be taken to be 2.4 kNm−2.

The following assumes that the load is carried by the inner leaf only. The load-bearing capacity of the wall istherefore based on the horizontal cross-sectional area of the inner leaf alone, although the stiffening effect of theother leaf has been taken into account when calculating the slenderness ratio.

LOADINGCharacteristic permanent load gk

gk = roof load + self weight of wall

=

× ×+ × ×

( . ) ( . ) .

6 5 1 62

3 5 1 2 4

= 19.5 + 8.4 = 27.9 kN per m run of wall

Characteristic imposed load, qk

qk = roof load =

×=

( . ) . .

6 5 11 52

4 9 kN per m run of wall

Ultimate design load, NEd

NEd = 1.35gk + 1.5qk = 1.35 × 27.9 + 1.5 × 4.9 = 45 kN m−1 run of wall

DESIGN VERTICAL LOAD RESISTANCE OF WALLSlenderness ratio (SR)Concrete slab provides ‘enhanced’ resistance to wall:

hef = 0.75 × height = 0.75 × 3500 = 2625 mm

t t tef mm . . = + = + =13

233 3 33 102 5 100 127 6

SR ef

ef

.

. = = = <ht

2625127 6

20 6 27

Fig. 10.5

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451

Capacity reduction factorEccentricity at the top and bottom of the wall, eiThe load from the concrete roof will be applied eccentrically as shown in the figure.The eccentricity is given by:

e

t t ttx .= − = =

2 3 60 167



M i/N = ex = t/6 = 100/6 = 16.7 mm


e init = hef/450 = 2625/450 = 5.83 mm

ei = M i/N i + ehi ± e init ≥ 0.05t

= 16.67 + 0 + 5.83 = 22.5 mm = 0.225t

Hence ei = 0.225t and Φi = 1 − 2e i/t = 1 − 2 × (0.225t)/t = 0.55


Mmd/Nmd ≈ 1/2 × eccentricity at top of wall = 1/2 × 16.7 = 8.35 mm

ehm = 0, since there are no horizontal loads

e init = hef/450 = 2625/450 = 5.83 mm

em = Mmd/Nmd + ehm ± einit ≥ 0.05t

= 8.35 + 0 + 5.83 = 14.18 mm = 0.1418t

emk = em + e k = 0.1418t + 0 = 0.1418t

Hence emk = 0.1418t

Modulus of elasticity of masonry, E = 1000fk (Cl. 2.9 of National Annex to EC 6)

u

ht

et

tt

.

( . )

.=−

−=

−

−=

ef

ef

mk

2

23 37

20 6 2

23 370 1418

1 05

A

et

tt1 1 2 1 2

0 14180 716

. .= − = − × =mk

Φm . ..

= = =− −A e e

u

12

1 05

2

2 2

0 716 0 41

Hence capacity reduction factor is 0.41

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452

Example 10.4 continuedDESIGN RESISTANCE OF WALL (NRd)Modification factorSmall plan area modification factor does not apply since horizontal cross-sectional area of wall, A = 6 × 0.1 = 0.6 m2

> 0.1m2.

Safety factor for materials (γγγγγM)Since manufacturing control of the units is Category II and the execution control is Class 1, from Table 10.6γM = 2.6

Design resistance

N

tf fRd

i,m k

M

k .

.= =

×Φγ

0 41 1002 6

DETERMINATION OF fk


NRd ≥ NEd

0.41 100 2.6

k× ×≥

f 45

Hence

fk

. ≥ =

45

15 772.85 Nmm−2

SELECTION OF BLOCK AND MORTAR TYPEAssuming the wall is made of Group 1 concrete blocks of length 440 mm, width 100 mm, height 215 mm laid in M4general purpose mortar

⇒ K = 0.55 (Table 10.5)

⇒ fm = 4 Nmm−2 (Table 10.4)

⇒ shape factor = 1.38

⇒ conditioning factor = 1.0 (assuming blocks are manufactured to EN 771-3)


f k = Kf b0.7f m

0.3

2.85 = 0.55 × f b0.7 × 40.3

Therefore the normalised compressive strength of the units, fb ≥ 5.8 Nmm−2.

The declared mean compressive strength of the units is given by

fb = conditioning factor × shape factor × declared mean compressive strength5.8 = 1.0 × 1.38 × declared mean compressive strength

Therefore the required declared air dried mean compressive strength of units ≥ 4.2 Nmm−2

The bricks and mortar for the outer leaf would be selected on the basis of appearance and durability.

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Example 10.5 Block wall subject to axial load and wind (EC 6)The inner leaf of the cavity wall shown in Fig. 10.6(a) is 6 m in length and required to resist a concentric axial loadof 135 kN m−1 run, 35 kN m−1 run at t/6 eccentricity at the top of the wall and a lateral load of 1.2 kN m−2. Assumingthe inner leaf is made of Group 1 solid concrete blocks of length 390 mm, width 100 mm and height 190mm, meanair dried compressive strength 20 Nmm−2, laid in mortar designation (iii), check the suitability of the design. Assumethe manufacturing control of the units is Category II and the execution control is Class 2.


LOADINGUltimate design load, NEd

NEd = 135 + 35 = 170 kN m−1 run of wall

DESIGN VERTICAL LOAD RESISTANCE OF WALLSlenderness ratio (SR)Concrete slab provides ‘enhanced’ resistance to wall:

hef = ρnh = 0.75 × 2800 = 2100 mm

t t tef mm . . = + = + =13

233 3 33 102 5 100 127 6

SR ef

ef

.

. = = = <ht

2100127 6

16 5 27

CAPACITY REDUCTION FACTOR

Eccentricity at the top of wallMid/Nid = 35 × (t/6)/N = 35 × (100/6)/170 = 3.43 mm

ehe = (WL/10)/N = (1.2 × 3 × 2800)/(10 × 170) = 5.93 mm (Fig. 10.6b)

einit = hef/450 = 2100/450 = 4.67 mm

The eccentricity of the design vertical load at the top and bottom of the wall, ei, is given by

ei = M i/N i + ehi ± einit ≥ 0.05t

= 3.43 + 5.93 + 4.67 = 14.03 mm = 0.14t

Hence ei = 0.14t and the capacity reduction factor, Φi = 1 − 2ei/t = 1 − 2 × (0.14t)/t = 0.72

(a) (b)

WL/10

WL/40h = 3000 mm

Fig. 10.6

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454

Eccentricity in middle of the wallMmd/Nmd ≈ 1/2 × eccentricity at top of wall = 1/2 × 3.43 = 1.72 mm

ehm = (WL/40)/N = (γFWk)L /40N = (1.2 × 3 × 2800)/(40 × 170) = 1.48 mm (Fig. 10.6b)

e init = hef/450 = 2100/450 = 4.67 mm

Eccentricity of the design vertical load in the middle of the wall, em, is given by

em = Mmd/Nmd + ehm ± einit ≥ 0.05t

= 1.72 + 1.48 + 4.67 = 7.87 mm = 0.0787t

Total eccentricity in the middle of the wall, emk = em + e k = 0.0787t + 0 = 0.0787t

Modulus of elasticity of masonry, E = 1000fk (Cl. 2.9 of National Annex)

u

ht

et

tt

.

( . )

.=−

−=

−

−=

ef

ef

mk

2

23 37

16 5 2

23 370 0787

0 722

A

et

tt1 1 2 1 2

0 07870 843

. .= − = − × =mk


= = × =− −A e e

u

12

0 722

2

2 2

0 843 0 65

DESIGN VERTICAL LOAD RESISTANCE OF WALL (NRd)Characteristic compressive strengthThe wall is made of Group 1 concrete blocks of length 390 mm, height 190 mm, thickness 100mm, mean compressivestrength 20 Nmm−2, laid in M4 general purpose mortar

⇒ K = 0.55 (Table 10.5)

⇒ fm = 4 Nmm−2 (Table 10.4)

⇒ shape factor = 1.32 (Table A1, EN 772–1)

⇒ conditioning factor = 1.0

The normalised mean compressive strength of the units, fb, is given by

fb = conditioning factor × shape factor × declared mean compressive strength

= 1.0 × 1.32 × 20 = 26.4 Nmm−2


f k = Kf b0.7f m

0.3 = 0.55 × 26.40.7 × 40.3 = 8.2 Nmm−2

Modification factorSmall plan area modification factor does not apply since horizontal cross-sectional area of wall, A = 6 × 0.1 = 0.6 m2

> 0.1 m2.

Safety factor for materials (γγγγγM)Since manufacturing control is Category II and execution control is Class 2, from Table 10.6 γM = 3

Design resistance of wall

N

tfNRd

i,m k

MEdkNm run

. . . = =

× ×= >−Φ

γ0 65 100 8 2

3177 6 1 OK


9780415467193_C10 9/3/09, 3:13 PM454

455


10.11 Design of laterally loadedwall panels

Like BS 5628, the design of panel walls in EC 6 isbased on ‘yield line’ principles. The design methodfor laterally loaded panel walls given in EC 6is identical to that used in BS 5628 and involvesverifying the design value of the moment, MEd,does not exceed the design value of the moment ofresistance of the wall, MRd i.e.

MEd ≤ MRd (10.18)

Since failure may take place about two axes(Fig. 5.21), for a given panel, there will be twodesign moments and two corresponding momentsof resistance. The ultimate design moment per unitheight of a panel when the plane of bending isperpendicular to the bed joint, MEd2 (Fig. 5.22), isgiven by:

MEd2 = α2WkγF l2 (10.19)

The ultimate design moment per unit height ofa panel when the plane of bending is parallel to thebed joint, MEd1, is given by:

MEd1 = µα2WkγFl2 (10.20)

whereµ is the orthogonal ratioα2 is the bending moment coefficient taken

from (Annex E)γF is the partial factor for actions (Table 10.2)l is the length of the panel between supportsWk is the characteristic wind load per unit

area

The Tables in Annex E are identical to Table 8of BS 5628, part of which is reproduced as Table5.14 of this book.

The corresponding design moments of resistancewhen the plane of bending is perpendicular, MRd2,or parallel, MRd1, to the bed joint is given by equa-tions 10.21 and 10.22 respectively:

MRd2 = fxd2Z (10.21)

MRd1 = fxd1Z (10.22)

wherefxd1 design flexural strength parallel to the plane

of bending, f

fxd1

xk1

M

=γ

(10.23)

fxd2 design flexural strength perpendicular to the

plane of bending, f

fxd2

xk2

M

=γ

(10.24)

Z elastic section modulusfxk1 characteristic flexural strength parallel to

the plane of bending (Table 6 of NationalAnnex to EC 6)

fxk2 characteristic flexural strength perpendicularto the plane of bending (Table 6 ofNational Annex to EC 6)

γM partial factor for materials (Table 10.6).

The characteristic flexural strength values givenin Table 6 of the National Annex to EC 6 areidentical to those presented in Table 3 of BS 5628,reproduced here as Table 5.13.

Equations 10.18–10.22 form the basis for thedesign of laterally loaded panel walls, ignoring anycontribution from self-weight and other verticalloads. It should be noted, however, since by defini-tion µ = fkx par /fkx perp, Mk par = µMk perp (by dividingequation 10.19 by equation 10.20) that either equa-tions 10.19 and 10.21 or equations 10.20 and 10.22can be used in design. The full design procedure issummarised in Fig. 5.27 and illustrated by meansof the following examples.

9780415467193_C10 9/3/09, 3:14 PM455


456

ULTIMATE DESIGN MOMENT (M)Since the vertical edges are unsupported, the panel must span vertically and, therefore, equations 10.19 and 10.20cannot be used to determine the design moment here. The critical plane of bending will be parallel to the bed jointand the ultimate design moment at mid-height of the panel, M, is given by

M =

×Ultimate load height8

Ultimate load on the panel = wind pressure × area

= (γFWk)(height × length of panel)

= 1.5Wk3000 × 1000 = 4.5Wk106 N m−1 length of wall

Hence

M

WW

. . =

× ×= ×

4 5 10 30008

1 6875 106

9kk Nmm m−1 length of wall

MOMENT OF RESISTANCE (Md)Section modulus (Z )

Z

bd

. . = =

×= ×

2 3 26 3

6

10 102 5


Moment of resistanceThe design moment of resistance, Md, is equal to the moment of resistance when the plane of bending is parallel tothe bed joint, Mk par. Hence

M M

f Zd k par

xk1

M

Nmm m . .

. . = = =

× ×= × −

γ0 5 1 75 10

2 70 324 10

66 1

where fxk1 = 0.5 N mm−2 from Table 5.13, since water absorption of clay bricks < 7% and the mortar is designation (iii).

DETERMINATION OF CHARACTERISTIC WIND PRESSURE (WK)For structural stability:

M ≤ Md

1.6875 × 109Wk ≤ 0.324 × 106

Wk ≤ 0.192 × 10−3 Nmm−2

Hence the characteristic wind pressure that the panel can resist is 0.192 × 10−3 Nmm−2 or 0.192 kNm−2.

Example 10.6 Analysis of a one-way spanning wall panel (EC 6)Estimate the characteristic wind pressure, Wk, that the cladding panel shown in Fig. 10.7 can resist assuming that itis constructed using clay bricks having a water absorption of < 7% and mortar designation (iii). Assume γF = 1.5 andγM = 2.7.

Fig. 10.7

9780415467193_C10 9/3/09, 3:14 PM456

457

Example 10.7 Analysis of a two-way spanning panel wall (EC 6)The panel wall shown in Fig. 10.7 is constructed using clay bricks having a water absorption of greater than 12 percent and mortar designation (ii). The clay bricks are Category II and the execution control is Class 2. Calculate thecharacteristic wind pressure, Wk, the wall can withstand assuming the wall is simply supported on all four edges.


µ

.

. . = = =

ff

xk1

xk2

0 30 9

0 33 (Table 5.13)

Therefore use µ = 0.35 values in Table 5.14 to determine the design bending moments acting on the wall.


hL

.= =34

0 75

Hence, α2 = 0.052 from Table 5.14 (E) (based on Annex E, Wall support condition E, EC 6)

Ultimate design momentM = Mperp = α2γFWkL2 = 0.052 × 1.5Wk × 42 kNm m−1 run = 1.248 × 106Wk Nmm m−1 run

Note that γF = 1.5 since wind load is the primary action.

MOMENT OF RESISTANCE (Md)Safety factor for materials (γγγγγm)Since the execution control is Class 2 γM = 2.7 (Table 10.6)

Section modulus (Z)

Z

bd

.= =

×2 3 2

610 102 5



M M

f Zd k perp

xk2

M

Nmm m run . .

. . = = =

× ×= × −

γ0 9 1 75 10

2 70 583 10

66 1

DETERMINATION OF CHARACTERISTIC WIND PRESSURE (WK)For structural stability

M ≤ Md

1.248 × 106Wk ≤ 0.583 × 106

⇒ Wk ≤ 0.467 kN m−2

Hence the panel is able to resist a characteristic wind pressure of 0.467 kN m−2.

Fig. 10.8


9780415467193_C10 9/3/09, 3:14 PM457

Eurocode 5: Design of timber structures

458

design values of actions and combination of actions,including values of the partial factors for actions,and EN 338 for material properties of timber suchas bending and shear strength, elastic modulus,density, etc. Reference may also have to be madeto Part 1.2 of Eurocode 5 (EN 1995–1–2) whichgives the general design rules for timber structuresin fire conditions. Like EC 5, Part 1.2 was pub-lished in final form in 2004 and the design methodsare essentially the same as those in the correspond-ing British Standard, BS 5268–4. Neither Part 1.2nor Part 2 of Eurocode 5 on timber bridges whichhas also been available in final form since 2004 arediscussed in this chapter.

11.2 LayoutIn common with the other structural Eurocodes,EC 5 has been drafted by a panel of experts drawnfrom the various EC Member State countries. It isbased on studies carried out by Working Commis-sion W18: Timber Structures of the CIB Interna-tional Council for Building Research Studies andDocumentation, in particular on CIB StructuralTimber Design Code: Report No. 66, published in1983. The following subjects are covered in EC 5:

Chapter 1: GeneralChapter 2: Basis of designChapter 3: Material propertiesChapter 4: DurabilityChapter 5: Basis of structural actionChapter 6: Ultimate limit statesChapter 7: Serviceability limit statesChapter 8: Connections with metal fastenersChapter 9: Components and assembliesChapter 10: Structural detailing and controlChapter 11: Special rules for diaphragm structures

Also included are Annexes for shear failure atconnections, for mechanically jointed beams and

This chapter briefly describes the content of Part 1.1 ofEurocode 5, the new European standard for the designof buildings in timber, which is expected to replace itsBritish equivalent BS 5268: Part 2 about 2010. Thechapter highlights the principal differences between theEuropean standard and BS 5268: Part 2. It alsoincludes a number of worked examples to illustrate thenew procedures for designing flexural and compressionmembers.

11.1 IntroductionEurocode 5 applies to the design of building andcivil engineering structures in timber. It is basedon limit state principles and comes in three Partsas shown in Table 11.1.

Part 1.1 of Eurocode 5, which is the subject ofthis discussion, gives the general design rules fortimber structures together with specific design rulesfor buildings. It is largely similar in scope to Part 2of BS 5268, which was discussed in Chapter 6.Part 1.1 of Eurocode 5, hereafter referred to asEC 5, was published as a preliminary standard,reference no: DD ENV 1995–1–1, in 1994 andthen in final form in 2004. It is expected to replaceBS 5268 about 2010.

Design of building structures cannot wholly beundertaken using EC 5, however. Reference willhave to be made to a number of other documents,notably EN 1990 and Eurocode 1 to determine the

Table 11.1 Scope of Eurocode 5: Design ofTimber Structures

Part Subject

1.1 General – Common Rules and Rules forBuildings

1.2 General rules – Structural Fire Design2 Bridges

Chapter 11

Eurocode 5: Designof timber structures

9780415467193_C11 9/3/09, 4:46 PM458

459

for built-up columns. All the Annexes are labelled‘informative’. As explained in section 7.5.4 of thisbook, this signifies that this material does nothave any status but has been included merely forinformation.

As can be appreciated from the above contentslist the organisation of material is different to thatused in BS 5268 but follows the layout adoptedin the other structural Eurocodes. Generally, thedesign rules in EC 5 are sequenced on the basis ofaction effects rather than on the type of member,as in BS 5268. The main reason cited for thischange in style is to avoid repetition of designrules and also to promote a better understandingof structural behaviour.

This chapter briefly describes the contents ofEC 5, in so far as it is relevant to the design offlexural and compression members in solid timber.The rules governing the design of joints or othertimber types are not discussed.

11.3 Principles/Application rulesAs with the other structural Eurocodes and for thereasons discussed in Chapter 7, the clauses in EC5 have been divided into Principles and Applica-tion rules. Principles comprise general statements,definitions, requirements and models for which noalternative is permitted. Principles are indicated bythe letter P after the clause numbers. The Appli-cation rules are generally recognised rules whichfollow the statements and satisfy the requirementsgiven in the Principles. The absence of the letter Pafter the clause number indicates an applicationrule. The use of alternative application rules tothose recommended in the Eurocode is permittedprovided it can be show that the alternatives are atleast equivalent and do not adversely affect otherdesign requirements. It is worth noting, however,that if an alternative application rule is used theresulting design will not be deemed Eurocodecompliant


Like the other structural Eurocodes, Eurocode 5allows some parameters and procedures to bedetermined at the national level. Where a nationalchoice is allowed this is indicated in a note inthe normative text under the relevant clause. Thenote may include the recommended value of the

parameter, or preferred method, etc., but the actualvalue, methodology, etc., to be used in a particularMember State country is given in the appropriateNational Annex. The recommended values of theseparameters and design methods/procedures arecollectively referred to as Nationally DeterminedParameters (NDPs). The NDPs determine variousaspects of design but perhaps most importantlythe level of safety of structures during execution(i.e. construction/fabrication) and in-service, whichremains the responsibility of individual nations.

The UK National Annex to EC 5 was publishedin 2006 and the discussion and worked examplesin this chapter are based on the material in EC 5and the NDPs in the accompanying NationalAnnex.

11.5 SymbolsChapter 1 of EC 5 lists the symbols used in EC 5.Those relevant to this discussion are reproducedbelow.

GEOMETRICAL PROPERTIES:b breadth of beamh depth of beamA areai radius of gyrationI second moment of areaWy,Wz elastic modulus about y–y (major

axis) and z–z (minor axis)

BENDING:l spanMd design momentG permanent actionQ variable actionσm,d design normal bending stressfm,k characteristic bending strengthfm,d design bending strengthγG partial coefficient for permanent

actionsγQ partial coefficient for variable actionsγM partial factor for material properties,

modelling uncertainties andgeometric variations

kmod modification factor to strength valuesk sys load sharing factork inst instability factor for lateral bucklingE0,05 fifth percentile value of modulus of

elasticityEmean mean value of modulus of elasticity

(parallel) to grain

Symbols

9780415467193_C11 9/3/09, 4:46 PM459


460

Gmean mean value of shear modulus =Emean/16

DEFLECTION:u inst instantaneous deformationu inst,G instantaneous deformation due to a

permanent action Gu inst,Q,1 instantaneous deformation for the

leading variable action Q1

ufin final deformationufin,G final deformation due to a permanent

action Gufin,Q,1 final deformation for the leading

variable action Q1

kdef deformation factorwnet,fin net final deflectionum bending deflectionuv shear deflection

VIBRATION:f1 fundamental frequency of vibrationb floor widthl floor lengthv unit impulse velocityζ damping coefficientn40 number of first order modes with

natural frequencies below 40 Hz

SHEAR:Vd design shear forceτd design shear stressfv,k characteristic shear strengthfv,d design shear strength

BEARING:F90,d design bearing forcel length of bearingσc,90,d design compression stress

perpendicular to grainfc,90,k characteristic compression strength

perpendicular to grainfc,90,d design compression strength

perpendicular to grain

COMPRESSION:lef effective length of columnλy,λ z slenderness ratios about y–y and z–z

axesλrel,y,λrel,z relative slenderness ratios about y–y

and z–z axesN design axial force

σc,0,d design compression stress parallel tograin

fc,0,k characteristic compression strengthparallel to grain

fc,0,d design compression strength parallelto grain

σm,y,d,σm,z,d design bending stresses parallel to grainfm,y,d, fm,z,d design bending strengths parallel to

grainkc compression factor

11.6 Basis of designAs pointed out above, EC 5, unlike BS 5268,is based on limit state principles. However, incommon with other limit state codes, EC 5 recom-mends that the two principal categories of limitstates to be considered in design are ultimate andserviceability limit states. A separate third limit stateof durability is also mentioned in section 4 of EC 5which covers the risk of timber decay due to fungalor insect attack as well as the risk of corrosion ofmetal fasteners and connections, e.g. nails, screwsand staples. Measures to reduce the risk of timberdecay include selecting materials which are natur-ally durable or the use of appropriate preservativetreatments. Possible measures against corrosionattack of metal fasteners include the use of zinccoatings or stainless steel.

The terms ultimate state and serviceability stateapply in the same way as is understood in otherlimit state codes. Thus ultimate limit states arethose associated with collapse or with other formsof structural failure which may endanger the safetyof people while serviceability limit states correspondto states beyond which specific service criteria areno longer met. The serviceability limit stateswhich must be checked in EC 5 are deflection andvibration. The ultimate limit states, which mustbe checked singly or in combination, include bend-ing, shear, compression and buckling. The variousdesign rules for checking these limit states arediscussed later.

11.6.1 ACTIONSAction is the Eurocode terminology for loads andimposed deformations. Permanent actions, G, areall the fixed loads acting on the structure, includingthe finishes, fixtures and self weight of the structure.Variable actions, Q, include the imposed, wind andsnow loads.

Clause 2.3.1.1 of EC 5 recommends thatthe actions to be used in design, principally

9780415467193_C11 9/3/09, 4:46 PM460

461

characteristic permanent, Gk, and variable, Qk,actions, should be taken from Eurocode 1: Actionson structures. Guidance on determining the designvalues of actions and combination of actions,including the partial safety factors for actions, aregiven in EN 1990: Basis of structural design. Thesedocuments and topics are briefly discussed in sec-tion 8.5 of this book. As noted there, the designvalue of an action (Fd) is obtained by multiplyingthe representative value (Frep) by the appropriatepartial safety factor for actions (γf):

Fd = γf Frep (11.1)

Table 11.2 shows the relevant partial safety fac-tors for the ultimate limit state of strength. Othersafety factors will apply in other design situations.For example, the partial factors for the ultimatelimit states of equilibrium are shown in Table 8.6.In equation 11.1, Frep is generally taken as the char-acteristic value of a permanent or variable action(i.e. Fk). Assuming that the member being designedis subjected to one or more permanent actions andone variable action only, i.e. load combination 1 inTable 11.2, the partial safety factor for permanentactions, γG, will normally be taken as 1.35 and forthe variable action, γQ as 1.5. As discussed in section8.5.3, it is possible to improve structural efficiencyby using expressions 6.10a and 6.10b of EN 1990(respectively, load combination 3(a) and 3(b)/3(c)in Table 11.2) to estimate the design values ofactions but the value of 1.35 for γG is conservativeand used throughout this chapter.

11.6.2 MATERIAL PROPERTIESEC 5, unlike BS 5268, does not contain the ma-terial properties, e.g. bending and shear strengths,






(a) 1.35 1.0 1.5ψ0,1 0 1.5ψ0,2

(b) 1.35/1.35ξ 1.0 1.5 0 1.5ψ0

(c) 1.35/1.35ξ 1.0 1.5ψ0 0 1.5

necessary for sizing members. This informationis to be found in a CEN supporting standard fortimber products, namely EN 338: Structural Timber:Strength classes.

Table 11.3 shows the range of timber strengthclasses available for design. In practice the mostcommonly recommended strength classes are C16and C24. The table also gives the characteristicstrength and stiffness properties and densityvalues for each class. Note that the strength classindicates the characteristic bending strength of thetimber. Comparison with the strength classes usedin BS 5268 (Table 6.3) shows that they are in factidentical except class TR26. However, there areconsiderable differences in the values of strengthfor the same class of timber. This is because thestrengths in EN 338 are fifth percentile valuesderived directly from laboratory tests of five min-utes duration whereas those in BS 5268 are gradestresses which have been reduced for long-termduration and already include a safety factor. Onebenefit of using characteristic values of materialproperties rather than grade stresses is that it willmake it easier to sanction the use of new materialsand component for structural purposes, since suchvalues can be utilised immediately, without firsthaving to determine what reduction factors areneeded to convert them to permissible or workingvalues.

The characteristic values of strength in Table 11.3are related to a depth in bending and width intension of 150 mm. For depths in bending or widthsin tension of solid timber, h, less than 150 mm thecharacteristic bending and tension strengths maybe increased by the factor kh given

Basis of design

9780415467193_C11 9/3/09, 4:46 PM461


462

Tab

le 1

1.3

Str

uctu

ral

tim

ber

stre

ngth

cla

sses

(T

able

1,

EN

338

)

Spe

cies

typ

eP

opla

r an

d co

nife

r sp

ecie

sD

ecid

uous

spe

cies

C14

C16

C18

C22

C24

C27

C30

C35

C40

D30

D35

D40

D50

D60

D70

Str

engt

h p

rop

erti

es (

Nm

m−−−− −2

)B

endi

ngf m

,k14

1618

2224

2730

3540

3035

4050

6070

Ten

sion

par

alle

lf t,

0,k

810

1113

1416

1821

2418

2124

3036

42T

ensi

on p

erpe

ndic

ular

f t,90

,k0.

30.

30.

30.

30.

40.

40.

40.

40.

40.

60.

60.

60.

60.

70.

9C

ompr

essi

on p

aral

lel

f c,0

,k16

1718

2021

2223

2526

2325

2629

3234

Com

pres

sion

f c,90

,k4.

34.

64.

85.

15.

35.

65.

76.

06.

38.

08.

48.

89.

710

.513

.5pe

rpen

dicu

lar

She

arf v,

k1.

71.

82.

02.

42.

52.

83.

03.

43.

83.

03.

43.

84.

65.

36.

0S

tiff

nes

s p

rop

erti

es (

kN m

m−−−− −

2)

Mea

n m

odul

us o

fE

0,m

ean

78

910

1112

1213

1410

1011

1417

20el

asti

city

par

alle

l5%

mod

ulus

of

E0,

054.

75.

46.

06.

77.

48.

08.

08.

79.

48.

08.

79.

411

.814

.36.

8el

asti

city

par

alle

lM

ean

mod

ulus

of

E90

,mea

n0.

230.

270.

300.

330.

370.

400.

400.

430.

470.

640.

690.

750.

931.

131.

33el

asti

city

per

pend

icul

arM

ean

shea

r m

odul

usG

mea

n0.

440.

500.

560.

630.

690.

750.

750.

810.

880.

600.

650.

700.

881.

061.

25D

ensi

ty (

kg m

−−−− −3 )

Den

sity

ρ k29

031

032

034

035

037

038

040

042

053

056

059

065

070

090

0A

vera

ge d

ensi

tyρ m

ean

350

370

380

410

420

450

460

480

500

640

670

700

780

840

1080

9780415467193_C11 9/3/09, 4:46 PM462

463

Basis of design

Table 11.5 Values of kmod for solid timber(based on Table 3.1, EC 5)

Load duration class Service class

1 2 3

Permanent 0.60 0.60 0.50Long-term 0.70 0.70 0.55Medium-term 0.80 0.80 0.65Short-term 0.90 0.90 0.70Instantaneous 1.10 1.10 0.90

Table 11.7 Load duration classes (Table 1 of National Annex toEC 5)

Load-duration class Order of accumulated duration Examples of loadingof characteristic load

Permanent > 10 years Self weightLong-term 6 months – 10 years StorageMedium-term 1 week – 6 months Imposed floor loadingShort-term < 1 week Snow, maintenanceInstantaneous Wind, impact loading

Table 11.4 Partial factors, γM, for solid timber(based on Table 2.3, EC 5)

Design situation γM

Fundamental combinations for solid timber 1.3Accidental combinations 1.0Serviceability limit states 1.0

by:

kh = lesser of

1500 2

h

.

and 1.3 (11.2)

The characteristic strengths, Xk, are convertedto design values, Xd, by dividing by a partial factor,γM, taken from Table 11.4, and multiplying by afactor kmod, obtained from Table 11.5.

Note that γM is not simply a partial factor formaterials but also takes account of modelling andgeometric uncertainties.

X k

Xd mod

k

M

=γ

(11.3)

EC 5, like BS 5268, allows the design strengthsdetermined using equation 11.3 to be multipliedby a number of other factors as appropriate suchas kcrit (section 11.4.4), kv (section 11.4.5), kc,90

(section 11.4.6) and the loading sharing factor, ksys,where several equally spaced similar members areable to resist a common load. Typical memberswhich fall into this category may include joists inflat roofs or floors with a maximum span of 6m andwall studs with a maximum height of 4m (clause5.4.6, ENV EC 5). According to clause 6.6 of EC5 a value of ksys = 1.1 may generally be assumed.

As can be seen from Table 11.5, kmod takes intoaccount the effect on strength parameters of theduration of load and the environmental conditionsthat the structure will experience in service. EC 5

Table 11.6 Service classes (based on clause2.3.1.3, EC 5)

Service Moisture Typical serviceClass content conditions

1 ≤ 12% 20°C, 65% RH, i.e. internal,heated conditions

2 ≤ 20% 20°C, 85% RH, i.e. internal butcold

3 > 20% Climatic conditions leading to ahigher moisture content than inservice class 2, i.e. external

defines three service classes, which are the same asthose in BS 5268, and five load duration classes,which differ from those in BS 5268, as summar-ised in Tables 11.6 and 11.7 respectively. See alsoTable 2 of National Annex to EC 5 for guidance onassignment of timber elements in building struc-tures to service class.

Where a load combination consists of actionsbelonging to different load duration classes the valueof kmod should correspond to the action with theshortest duration. For example, for a permanentload and medium-term combination, a value of

9780415467193_C11 9/3/09, 4:46 PM463


464

kmod corresponding to the medium-term load shouldbe used.

Having discussed these more general aspects itis now possible to describe in detail EC 5 rulesgoverning the design of flexural and compressionmembers.

11.7 Design of flexural membersThe design of flexural members principally involvesconsideration of the following actions which arediscussed next:

1. Bending2. Deflection3. Vibration4. Lateral buckling5. Shear6. Bearing

11.7.1 BENDING (CL. 6.1.6, EC 5)If members are not to fail in bending, the followingconditions should be satisfied:

σ σm,y,d

m,y,dm

m,z,d

m,z,dfk

f + ≤ 1 (11.4)

k

f fmm,y,d

m,y,d

m,z,d

m,z,d

σ σ + ≤ 1 (11.5)

whereσm,y,d and are the design bending stressesσm,z,d about axes y–y and z–z as shown in

Fig. 11.1fm,y,d and are the corresponding design bendingfm,z,d strengthskm is a factor that allows for the

redistribution of secondary bendingstresses and assumes the followingvalues:– for rectangular or square sections;

km = 0.7– for other cross-sections;

km = 1.0

Fig. 11.1 Beam axes.

Fig. 11.2 Components of deflection

Winst

Wcreep

l

Wc

Wnet,fin

Wfin

It should be noted that in EC 5 the x–x axis isthe axis along the member and that axes y–y andz–z are the major and minor axes respectively. Thesedefinitions are consistent with the other structuralEurocodes.

For beams with rectangular cross-sections

σm,y,d =y,d

y

y,d /

=M

W

M

bh2 6(11.6)

σm,z,d =z,d

z

z,d /

=M

W

M

hb2 6(11.7)

whereMy,d and Mz,d are the design bending moments

about y–y (major axis) and z–z(minor axis)

Wy and Wz are the values of elastic modulusabout y–y and z–z

b is the breadth of beamh is the depth of beam

11.7.2 DEFLECTION (CL. 7.2, EC 5)To prevent the possibility of damage to surfacingmaterials, ceilings, partitions and finishes, and tothe functional needs as well as aesthetic require-ments, EC 5 recommends various limiting valuesof deflection for beams (see Table 7.2, EC 5). Thecomponents of deflection are shown in Figure 11.2,where the symbols are defined as:

wc is the precamber (if applied)winst is the instantaneous deflection due to

permanent and variable actionswcreep is the creep deflection due to permanent

and variable actions

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465

Design of flexural members

Table 11.8 Limiting values for deflections of beams (based on Table 4 of National Annex to EC 5)

Type of member Deflection limits for individual beams, wfin

A member of span, A member withl between two supports a cantilever, l

Roof of floor members with a plastered l/250 l/125or plasterboard ceiling

Roof of floor members without a l/150 l/75plastered or plasterboard ceiling

Table 11.9 Values of kdef for solid timber to EN14081-1 and glue laminated timber to EN 14081(based on Table 3.2, EC 5)

Material Service class

1 2 3

Solid timber 0.60 0.80 2.00Glued laminated timber 0.60 0.80 2.00

wfin is the final deflection due to permanentand variable actions = winst + wcreep

wnet,fin is the net final deflection due to permanentand variable actions = wfin − wc

Note that if no precamber is applied net finaldeflection is equal to final deflection.

Clause 2.5 of the National Annex to EC 5reiterates the advice in clause 4.2(2) of Annex A1of EN 1990 that deflection and other serviceabilityrequirements should be specified for each projectand agreed with the client, but for guidance pro-vides the values for net final deflection of beamsin Table 11.8, which take into account creepdeformations.

According to clause 2.2.3(5) when the membersupports one or permanent actions but only a singlevariable action (i.e. Q1) the final deflection, ufin, isgiven by

ufin = ufin,G + ufin,Q1 (11.8)

where

ufin,G = u instG(1 + kdef) (11.9)

ufin,Q1 = u instQ1(1 + ψ2kdef) (11.10)

in whichkdef is the deformation factorψ2 is the factor for quasi permanent value of

permanent actions (Table 8.8)

The instantaneous deflections, i.e. u instG andu instQ1, can be calculated by means of the expres-sions given in Table 6.9 in this book and usingE0,mean or E90,mean as appropriate. Note that unlikeBS 5268 it is necessary to calculate the deflectionsproduced by permanent and variable actions sep-arately in EC 5. The final deflections are derivedfrom the instantaneous deflection via kdef whichtakes into account the combined effect of creepand moisture content. Recommended values of kdef

are given in Table 11.9.

11.7.3 VIBRATION (CL. 7.3, EC 5)EC 5, unlike BS 5268, gives procedures for calcu-lating the vibration characteristics of residentialfloors which must satisfy certain requirementsotherwise the vibrations may impair the functioningof the structure or cause unacceptable discomfortto users.

The fundamental frequency of vibration of arectangular residential floor, f1, can be estimatedusing the following expression and should normallyexceed 8 Hz

f

EIm1 22

( )

=πl

l (11.11)

wherem mass equal to the self weight of the floor

and other permanent actions per unit areain kg m−2

l floor span in m(EI )l equivalent bending stiffness in the beam

direction Nm2 m−1

For residential floors with a fundamental fre-quency greater than 8 Hz the following conditionsshould also be satisfied:

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466

s is the joist spacing in mmleq is the span, l, in mm, for simply

supported single span joistskamp 1.05 for simply supported solid timber

joists

The value of ν may be estimated from

ν

( . . )

=++

4 0 4 0 6200

40nmbl

in m/Ns2 (11.16)

whereb is the floor width in ml is the floor length in mn40 is the number of first order modes with

natural frequencies below 40 Hz and isgiven by

nf

b EIEI40

1

2 40 25

401

( )( )

.

=

−

l

l

b

(11.17)

where (EI )b is the equivalent plate bending stiffnessparallel to the beams and the other symbols are asdefined above.

11.7.4 LATERAL BUCKLING OF BEAMS(CL. 6.3.3, EC 5)

Where both lateral displacement of the compressionedge throughout the length of the member andtwisting of the member at supports are prevented,lateral buckling should not occur. Otherwise themember may be vulnerable to lateral buckling andthe rules in Cl. 6.3.3 (3) of EC 5 should be usedto assess the bending behaviour. Generally, the fol-lowing condition should be verified

σm,d ≤ kcrit fm,d (11.18)

whereσm,d design bending stressfm,d design bending strengthkcrit is a factor which takes into account the

reduction in bending strength due to lateralbuckling and is given by

kcrit = 1 for λ rel,m ≤ 0.75 (11.19)

kcrit = 1.56 − 0.75λrel,m

for 0.75 < λ rel,m ≤ 1.4 (11.20)

kcrit = 1/λ2rel,m for 1.4 < λ rel,m (11.21)

where λ rel,m is the relative slenderness ratio forbending given by

λ

σrel,mm,k

m,crit

=f

(11.22)

Table 11.10 Values of the parameters a and b(based on Table 5 of National Annex to EC 5)

Parameter Limit

a 1.8 mm for l ≤ 4 000 mm16 500/l1.1 mm for l > 4 000 mmWhere l is the joist span in mm

b For a ≤ 1 mm b = 180 − 60aFor a > 1 mm b = 160 − 40a

w

Fa ≤ −mmkN 1 (11.12)

and

ν ≤ b( f1ζ−1) m/(Ns−2) (11.13)

where

ζ is the modal damping coefficient, normallytaken as 0.02

w is the maximum vertical deflection caused bya concentrated static force F = 1.0

v is the unit impulse velocitya is the deflection of floor under a 1 kN point

load obtained from Table 11.10b is the velocity response constant obtained

from Table 11.10.

The value of w in equation 11.12 may be esti-mated from the following expression given in clause2.6.2 of the National Annex to EC 5

w

k kEI

=1000

48dist eq

3amp

joist( )

l(11.14)

wherekdist is the proportion of point load acting on

a single joistleq is the equivalent floor span in mmkamp is the amplification factor to account for

shear deflection in the case of solidtimber

(EI )joist is the bending stiffness of a joist inNmm2 (calculated using Emean)

kdist = kstrut[0.38 − 0.08 ln[14(EI)b/s4] ≥ 0.30

(11.15)

in whichkstrut = 0.97 for appropriately installed single

or multiple lines of strutting, orotherwise 1.0

(EI )b is the floor flexural rigidity perpendicularto the joists in Nmm2 m−1

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467


Table 11.11 Effective length as a ratio of the span (based onTable 6.1, EC 5)

Beam type Loading type lef/la

Simply supported Constant moment 1.0Uniformly distributed load 0.9Concentrated force at the middle of the span 0.8

Cantilever Uniformly distributed load 0.5Concentrated force at the free end 0.8

aThe ratios are valid for beams with torsionally restrained supports, loaded at thecentre of gravity. If the load is applied at the compression edge of the beam, lef shouldbe increased by 2 h and may be decreased by 0.5 h for a load at the tension edge ofthe beam.

τd

Ed .

=1 5V

A(11.26)

whereVEd is the design shear forceA is the cross-sectional area

The design shear strength, fv,d, is given by

f

k fv,d

mod v,k

M

=γ

(11.27)

where fv,k is the characteristic shear strength (Table11.3)

For beams notched at their ends as shown inFigure 11.3, the following condition should bechecked

τd ≤ kv.fv,d (11.28)

where

τd

Ed

ef

.

=1 5V

bh(see Fig. 11.3) (11.29)

kv is the shear factor which may attain the fol-lowing values:

For beams notched at the oppostie side tosupport (Fig. 11.3b)

kv = 1

For beams of solid timber notched at the sameside as support (Fig. 11.3a)

k

ki

h

hxh

v

n

.

( ) .

.

≤+

− + −

≤1

1 1

1 0 81

1

1 5

2α αα

α(11.30)

wherefm,k is the characteristic bending stressσm,crit is the critical bending stress generally

given by

σ

πm,crit

y,crit

y

z tor

ef y

, ,= =MW

E I G I

W0 05 0 05

l(11.23)

and for softwoods with solid rectangular sectionsshould be taken as

σm,crit

ef

.

,=0 78 2

0 05b

hE

l(11.24)

wherelef is the effective length of the beam, according

to Table 11.11b width of beamh depth of beamIz is the second moment of area about z-zItor is the torsional moment of inertiafm,k characteristic bending strengthE0,05 is the fifth percentile modulus of elasticity

parallel to grain (Table 11.3)G0,05 is the fifth percentile shear modulus

= E0,mean/16

11.7.5 SHEAR (CL. 6.1.7 & 6.5, EC 5)If flexural members are not to fail in shear, thefollowing condition should be satisfied:

τd ≤ fv,d (11.25)

whereτd is the design shear stressfv,d is the design shear strength (Table 11.3)

For a beam with a rectangular cross-section, thedesign shear stress occurs at the neutral axis and isgiven by:

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468

hef

hef

i(h–hef)(h–hef)

Fig. 11.3 End notched beams (Fig. 6.11, EC 5).

Fig. 11.4 Compression perpendicular to grain (based on Fig. 6.2, EC 5).

lla

h

b

wherekn = 5 for solid timberi is the notched inclination as defined in

Fig. 11.3h is the beam depth in mmx is the distance from line of action to the corner

α =

hhef (see Fig. 11.3) (11.31)

11.7.6 COMPRESSION PERPENDICULAR TO THEGRAIN (CL. 6.1.5, EC 5)

For compression perpendicular to the grain thefollowing condition should be satisfied:

σc,90,d ≤ kc,90 fc,90,d (11.32)

whereσc,90,d is the design compressive stress

perpendicular to grainfc,90,d is the design compressive strength

perpendicular to grainkc,90 is the compressive strength factor.

The factor kc,90 principally takes into account theeffect of support position and bearing length onbearing strength. For example in the case of a beamb wide and h deep, resting on end and internalsupports, bearing length l and overhang a ≤ h/3(Fig. 11.4), kc,90 at the end support is given by

k

hc, . .90 2 38

2501

124 0= −

+

≤

l

l(11.33)

and at internal supports is given by

k

hc, . .90 2 38

2501

124 0= −

+

≤

l

l(11.34)

Clause 6.1.5(4) of EC 5 gives details of other‘member arrangements’ and associated expressionsfor kc,90. In all cases the higher value of kc,90 willapply but with an upper limit of 4. If none of themember arrangements are appropriate, however,the value of kc,90 should be taken as 1.

It should be noted that there is a proposedamendment to this check which is due to be imple-mented sometime after publication of this book.Like the present method it involves checking thatthe following condition is satisfied

σc,90,d ≤ kc,90 fc,90,d

but with

σc,90,d =

F

Ac,90,d

ef

where Aef is the effective contact area in compres-sion perpendicular to the grain and the other sym-bols are as defined above.

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469

l

l1 l

a

h

b


Fig. 11.5 Member on discrete support (based on Fig. 6.2b, prEN 1995-1-1: A1: 2007).

members on discrete supports, provided l1 ≥ 2h,the value of kc,90 may be taken as 1.5 for solidsupport softwood timber. Other values apply formembers on continuous supports and for gluedlaminated softwood timber.

According to the amended clause 6.1.5, Aef

should be determined taking into account an effec-tive contact length parallel to the grain, where thecontact length, l, at each side is increased by 30 mmbut not more than a, l or l1/2 (Fig. 11.5). Thevalue of kc,90 should generally be taken as 1.0. For

DESIGN ACTIONSPermanent action, GkTongue & grove boarding = 0.10 kN m−2

Ceiling = 0.20 kN m−2

Joists (say) = 0.10 kN m−2

Total characteristic permanent action = 0.40 kN m−2

Example 11.1 Design of timber floor joists (EC 5)Design the timber floor joists for a domestic dwelling using timber of strength class C16 given that the:

a) floor width, b, is 3.6 m and floor span, l, is 3.4 mb) joists are spaced at 600 mm centresc) flooring is tongue and grove boarding of thickness 21 mm and a self-weight of 0.1 kN m−2

d) ceiling is of plasterboard with a self weight of 0.2 kN m−2

e) the bearing length is 100 mm.

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470

Variable action, QkImposed floor load for domestic dwelling (Table 3 of UK National Annex to EN 1991–1–1) is 1.50 kN m−2

Design actionTotal design load is

γGGk + γQQk = 1.35 × 0.40 + 1.5 × 1.5 = 2.79 kN m−2

Design load/joist, Fd = joist spacing × effective span × load

= 0.6 × 3.4 × 2.79 = 5.7 kN

CHARACTERISTIC STRENGTHS AND MODULUS OF ELASTICITY FOR TIMBER OF STRENGTHCLASS C22Values in N mm−2 are given as follows:

Bending Compression Shear parallel Modulus ofstrength perpendicular to grain to grain elasticity(fm,k) (fc,90,k) (fv,k) (E0,mean)

22.0 5.1 2.4 10000

BENDINGBending moment

M

Fd,y

d kNm . .

. = =×

=l

8

5 7 3 4

82 42

Assuming that the average moisture content of the timber joists does not exceed 20 per cent during the life of thestructure, design the joists for service class 2 (Table 11.6). Further, since the joists are required to carry permanentand variable (imposed) actions, the critical load duration class is ‘medium-term’ (Table 11.7). Hence from Table 11.5,kmod equals 0.8. From Table 11.4, γM (for ultimate limit states) = 1.3. Since the joists form part of a load sharingsystem the design strengths can be multiplied by a load sharing factor, ksys = 1.1.

Assuming kh = 1, design bending strength about the y-y axis is

f k k k

fm,y,d h sys mod

m,k

M

N mm . . . .

. = = × × × = −

γ1 0 1 1 0 8

22

1 314 9 2

The design bending stress is obtained from equation 11.4. Hence

σ σm,y,d

m,y,dm

m,z,d

m,z,dfk

f + ≤ 1

σm,y,d

m,z,d14 90 7

01

. . + × ≤

f

⇒ σm,y,d ≤ 14.9 N mm−2

W req

My

d,y

m,y,d

mm .

. ≥ =

×= ×

σ2 42 10

14 9162 10

63 3


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471

From Table 6.8 a 63 mm × 200 mm joist would be suitable (Wy = 420 × 103 mm3, Iy = 42 × 106 mm4, A =12.6 × 103 mm2).

Since h > 150 mm, kh = 1 (as assumed).

DEFLECTION

Instantaneous deflection due to permanent actions, uinstGFrom Table 11.4, γG for serviceability limit states = 1.0 and factored permanent load, G = γGGk = 1.0 × 0.40 =0.40 kN m−2

Factored permanent load per joist, Fd,G, is

Fd,G = total load × joist spacing × span length = 0.40 × 0.6 × 3.4 = 0.82 kN

From Table 6.9, instantaneous deflection due to permanent actions, uinstG, is given by

uinstG = bending deflection + shear deflection

= × + ×

5384

125

3F LEI

F LEA

d d

= ×

× ×× × ×

+ ×× × ×

× × ×=

. ( . )

. .

.

5

384

0 82 10 3 4 10

10 10 42 10

12

5

0 82 10 3 4 10

10 10 12 6 10

3 3 3

3 6

3 3

3 31 mm

Instantaneous deflection due to variable action, uinstQFrom Table 11.4, γQ for serviceability limit state = 1.0 and factored variable action, Q = γQQk = 1.0 × 1.5 =1.5 kN m−2

Factored variable action per joist, Fd,Q, is

Fd,Q = total load × joist spacing × span length = 1.5 × 0.6 × 3.4 = 3.06 kN

From Table 6.9, instantaneous deflection due to variable load, uinstQ, is given by

uinstQ = bending deflection + shear deflection

= × + ×

5384

125

3F LEI

F LEA

d,Q d,Q

= ×

× ×× × ×

+ ×× × ×

× × ×= + =

. ( . )

. .

. . . .

5

384

3 06 10 3 4 10

10 10 42 10

12

5

3 06 10 3 4 10

10 10 12 6 103 7 0 2 3 9

3 3 3

3 6

3 3

3 3mm

Final deflection due to permanent actionsFrom Table 11.9, for solid timber members subject to service class 2 and medium-term loading, Kdef = 0.8. Finaldeflection due to permanent actions, ufin,G, is given by

ufin,G = uinstG(1 + kdef) = 1 × (1 + 0.8) = 1.8 mm

Final deflection due to variable actionFrom Table 8.8, ψ2 = 0.3. Hence final deflection due to variable action, ufin,Q1, is given by

ufin,Q1 = uinstQ1(1 + ψ2kdef) = 3.9 × (1 + 0.3 × 0.8) = 4.9 mm



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472

Example 11.1 continuedCheck final deflectionTotal final deflection, ufin = ufin,G + ufin,Q1 = 1.8 + 4.9 = 6.7 mmPermissible final deflection (assuming the floor supports brittle finishes), wfin, is

wfin = 1/250 × span

= 1/250 × 3.4 × 103

= 13.6 mm > 6.7 mm OK

Therefore 63 mm × 200 mm joists are adequate in deflection

VIBRATIONAssuming that f1 > 8 Hz, check that

wF

a ≤ and ν ≤ b(f1ζ−1)

Check w/F ratiok strut = 1.0

(EI)b = 10 × 103 × (1000 × 213/12) = 7.72 × 109 Nmm2 m−1

Joist spacing, s = 600 mm

Hence kdist is

kdist = k strut[0.38 − 0.08 ln[14(EI)b/s4] ≥ 0.30

= 1.0 × [0.38 − 0.08 ln[14 × 7.72 × 109/6004] = 0.4

leq = 3400 mm

kamp = 1.05

(EI)joist = 10 × 103 × (63 × 2003/12) = 4.2 × 1011 Nmm2 m−1

Therefore the maximum vertical deflection caused by a concentrated static force F = 1.0, w, is

w

k k

EIa

. .

. . . = =

× × ×× ×

= < =1000

48

1000 0 4 3400 1 05

48 4 2 100 82 1 8

3

11

dist eq3

amp

joist( )mm mm

lOK

Check impulse velocityFloor width, b = 3.4 m and floor span, l = 3.6 m.

Iy is the second moment of area of the joist (ignore tongue and groove boarding unless a specific shear calculation atthe interface of joist and board is made):

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473


Example 11.1 continuedIy = 42 × 106 mm4 = 42 × 10−6 m4

E0,mean = 10 000 N mm−2 = 10 × 109 N m−2 (Table 11.3)

(EI)l = E0,meanIy/joist spacing

= 10 × 109 × 42 × 10−6/0.6 = 7 × 105 Nm2 m−1

Mass due to permanent actions per unit area, m, is

m = permanent action/gravitational constant

= 0.40 × 103/9.8 = 40.8 kg m−2

The fundamental frequency of vibration, f1, is

f

EIm1 2 2

5

2 2 3 47 10

40 8

( )

.

. = =

××

=π πl

l 17.8 Hz > 8 Hz as assumed

The number of first order modes, n40, is

n

f

b EI

EI40

1

2 4 0 25 2 4 3

3

0

401

40

17 81

3 4

3 6

700 10

7 72 10

( )

( )

.

.

.

.

.

=

−

=

−

××

l

l

b

..

.

25

4 13=

The unit impulse velocity, v, is

ν

( . . )

( . . . )

. . . . =

++

=+ ×

× × += − −4 0 4 0 6

200

4 0 4 0 6 4 13

40 8 3 4 3 6 2000 01640 1 2n

mbs

lm N

Assume damping coefficient, ζ = 0.02. From Table 11.10 since a (= 0.82 mm) < 1, b, is given by

b = 180 − 60a = 180 − 60 × 0.82 = 131

Hence permissible floor velocity = b(f.ζ−1) = 131(17.8 × 0.02 − 1)

= 0.04 m N−1 s−2 > 0.016 m N−1 s−2 OK

LATERAL BUCKLINGThis check is unnecessary as the compressive edge cannot move laterally because the joists are attached to tongueand groove boarding.

SHEARDesign shear strength is

f k k

fv,d sys mod

y,k

M

N mm . . .

. . = = × × = −

γ1 1 0 8

2 4

1 31 62 2


V

FEd

d

2N

. . = =

×= ×

5 7 10

22 85 10

33

Design shear stress at neutral axis is

τd

Edv,dNmm

.

. .

. . = =

× ××

= <−1 5 1 5 2 85 10

12 6 100 34

3

32V

Af OK

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474

Example 11.1 continuedBEARINGDesign compressive stressDesign bearing force is

F90,d = Fd/2 = 5.7 × 103/2 = 2.85 × 103 N

Assuming that the floor joists span onto 100 mm wide walls as shown above, the bearing stress is given by:

σc,90,d

d Nmm .

. ,= =

××

= −F

b90

322 85 10

63 1000 45

l

Design compressive strengthDesign compressive strength perpendicular to grain, fc,90,d, is given by

f k k

kc,90,d sys mod

c,90,k

M

N mm . . .

. . = = × × = −

γ1 1 0 8

5 1

1 33 4 2

Bearing capacityBy comparing the above diagram with Fig. 11.4, it can be seen that

a = 0, l = 100 mm and h = 200 mm

Since a < h/3

k

hc,90 . .

. = −

+

= −

+×

= <2 38

2501

122 38

100250

1200

12 1002 31 4

l

lOK

kc,90fc,90,d = 2.31 × 3.4 = 7.9 Nmm−2 > σc,90,d OK

Alternatively, using the procedure in clause 6.1.5 of prEN 1995-1-1: A1: 2007 gives l1 = a = 0 (by comparing theabove diagram with Fig. 11.5). Therefore, Aef = bl and σc,90,d = 0.45 Nmm−2 as before. Since l1 = 2h, kc,90 = 1.00. Hence

kc,90fc,90,d = 1.0 × 3.4 = 3.4 Nmm−2 > σc,90,d OK

CHECK ASSUMED SELF-WEIGHT OF JOISTSFrom Table 11.3, density of timber of strength class C22 is 410 kg/m3. Hence, self-weight of the joists, SW, is

SW

kg mkN m assumed

.

. . =

× × × × ×= <

− − −−63 200 10 410 9 8 10

0 60 084

6 3 32 OK

Finally, it is worth noting that C22 is not a preferred strength class and readers may wish to redesign the joists usingthe more commonly specified timber grades namely C16 or C24.

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475


Example 11.2 Design of a notched floor joist (EC 5)The joists in Example 11.1 are to be notched at the bearings with a 75m deep notch as shown below. Check thenotched section is still adequate.

The presence of the notch only affects the shear stress in the joists.

FACTOR K v

For beams notched at the loaded side, k v is taken as the lesser of 1 and the value calculated using equation 11.30.Comparing the above diagram with Fig. 11.3 gives

i = 2x = 75 mmα = he/h = 125/200 = 0.625

From clause 6.5.2(2) kn = 5 for solid timber. Therefore kv is

kk

ih

hxh

v

n

.

( ) .

.

≤+

− + −

≤1

1 1

1 0 81

1

1 5

2α αα

α

=× +

×

− + × −

=

.

. ( . ) . .

.

.

.

5 11 1 2

200

200 0 6251 0 625 0 875200

10 625

0 625

0 53

1 5

2

The shear strength, fv,d, is

f k k

fv,d sys mod

v,k

M

N mm . . .

. . = = × × = −

γ1 1 0 8

2 4

1 31 62 2

For a notched member, the design shear strength is given by

kv.fv,d = 0.53 × 1.62 = 0.86 N mm−2

Design shear stress, τd =

1 5 1 5 2 85 1063 125

3.

. .

Vbh

Ed

e

=× ×

× = 0.54 Nmm−2 < 0.86 N mm−2 OK

Therefore the section is also adequate when notched with a 75 mm deep bottom edge notch at the bearing.

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476

Example 11.3 Analysis of a solid timber beam restrained at supports(EC 5)

A solid timber beam, 75 mm wide × 250 mm deep, in strength class C16, 3.4 m simply supported, supports uniformlydistributed permanent (including self-weight of beam) and variable actions of respectively 0.2 kNm−1 and 2 kNm−1.Assuming the beam is torsionally restrained at supports and the exposure is service class 2 check its bending capacity.

X

X

h = 250 mm

b = 75 mm

Section X-X

DESIGN ACTIONTotal design load is

Fd = γGGk + γQQ k = 1.35 × 0.20 + 1.5 × 2 = 3.27 kN m−1

DESIGN BENDING STRESS

Bending moment

M

Fy,d

d kNm . .

. = =×

=l

8

3 27 3 4

84 73

2

σm,y,d

y,d

y

Nmm .

/ . = =

××

= −M

Z

4 73 10

75 200 69 5

6

22

Bending strengthAs the compressive edge of the beam is not restrained the bending strength is

= kcritfm,d

Since the load is uniformly distributed and applied to the top of the beam, from Table 11.11, lef = 0.9l + 2h = 0.9 ×3500 + 2 × 200 = 3550 mm

σm,crit

ef

.

.

. .,= =

××

× × =0 78 0 78 75

200 35505 4 10 33 37

2

0 05

23b

hE

l

λ

σrel,mm,k

m,crit

.

.= = =f 16

33 370 69

Since λrel,m < 0.75, kcrit = 1

k f k k k

fcrit m,y,d crit h mod

m,k

Mm,d

-2N mm Nmm . . . .

. ( . )=

= × × ×

= > =−

γσ1 0 1 0 0 8

16

1 39 85 9 52 OK

9780415467193_C11 9/3/09, 4:50 PM476

477

11.8 Design of columnsColumns are normally subjected to either axialload or combined axial load and bending. Axiallyloaded members may fail in compression or flexuralbuckling depending upon the relative slendernessratios, λrel,z and λ rel,y. Members subject to axial loadand bending are also susceptible to these modes offailure but may additionally fail due to lateral tor-sional buckling. The following subsections discussthe rules relevant to the design of members subjectto these two types of stress states.

11.8.1 COLUMNS SUBJECT TO AXIAL LOADONLY (CL. 6.3.2, EC 5)

The relative slenderness ratios λ rel,z and λ rel,y aredefined in EC 5 as follows

λ

λπrel,y

y c,0,k

0,05

=fE

(11.35)

and

λ

λπrel,z

z c,0,k

0,05

=fE

(11.36)

whereλy and λrel,y are slenderness ratios corresponding

to bending about the y-axis(deflection in the z-direction)

λz and λrel,z are slenderness ratios correspondingto bending about the z-axis(deflection in the y-direction)

E0,05 is the fifth percentile value of themodulus of elasticity parallel to thegrain.

The slenderness ratio, λ, is given by:

λ =

lef

i(11.37)

wherelef is the effective lengthi is the radius of gyration

EC 5 does not include a method for determiningthe effective length of a column. Therefore, design-ers will have to refer to the recommendation in BS5268: Part 2, as discussed in section 6.7.1 of thisbook.

Where both λrel,z and λrel,y are less than or equalto 0.3, columns subject to axial load only shouldsatisfy the following condition:

σc,0,d ≤ fc,0,d (11.38)

wherefc,0,d is the design compressive strength obtained

from eq. 11.3σc,0,d is the design compressive stress given by

σc,0,d =

NA

(11.39)

in whichN is the axial loadA is the cross-sectional area

In cases where either λ rel,z or λ rel,y exceeds 0.3,the column should satisfy the more stringent of thefollowing:

σc,0,d ≤ kc,y fc,0,d (11.40)

σc,0,d ≤ kc,z fc,0,d (11.41)

where

k

k kc,y

y y rel,y

( )

=+ −

12 2λ

(11.42)

k

k kc,z

z z rel,z

( )

=+ −

12 2λ

(11.43)

in which

ky = 0.5(1 + βc(λ rel,y − 0.3) + λ2rel,y) (11.44)

k z = 0.5(1 + βc(λ rel,z − 0.3) + λ2rel,z) (11.45)

whereβc = 0.2 (for solid timber).

11.8.2 COLUMNS SUBJECT TO AXIAL LOAD ANDBENDING (CL. 6.3.2, EC 5)

In this case if the relative slenderness ratios aboutboth the y–y and z–z axis of the column, λ rel,y andλrel,z respectively, are less than or equal to 0.3 thesuitability of the section can be assessed using themore stringent of the following conditions

σ σ σc,0,d

c,0,d

m,y,d

m,y,dm

m,z,d

m,z,df fk

f

+ + ≤

2

1 (11.46)

σ σ σc,0,d

c,0,dm

m,y,d

m,y,d

m,z,d

m,z,dfk

f f

+ + ≤

2

1 (11.47)

whereσc,0,d is the design compressive stress from

eq. 11.39fc,0,d is the design compressive strength from

eq. 11.3km = 0.7 for rectangular sections and

= 1.0 for other cross-sections

9780415467193_C11 11/3/09, 11:19 AM477


478

Design of columns

moment about the major axis (y–y) act, the columnmay be susceptible to lateral torsional buckling.The risk of this mode of failure occurring can beassessed using the following expression taken fromclause 6.3.3(6):

σ σm,d

crit m,d

c,d

c,z c,0,dk f k f

+ ≤

2

1 (11.50)

where σm,d is the design bending stress, σc,d is thedesign compressive strength parallel to the grain,kcrit is given by equations 11.19–11.21 and kc,z isgiven by equation 11.43.

Where either λrel,z or λrel,y exceeds 0.3, the col-umn is vulnerable to flexural buckling and the morestringent of the following expressions should besatisfied

σ σ σc,0,d

c,y c,0,d

m,y,d

m,y,dm

m,z,d

m,z,dk f fk

f + + ≤ 1 (11.48)

σ σ σc,0,d

c,z c,0,dm

m,y,d

m,y,d

m,z,d

m,z,dk fk

f f + + ≤ 1 (11.49)

Moreover in cases where λrel,z or λrel,y exceeds 0.3and when a combined compressive force and a

SLENDERNESS RATIO

λ λy z

ef = =l

ilef = 1.0 × h = 1.0 × 3750 = 3750 mm (see Table 6.11)

i

IA

dbbd

b

( / ) .= = = = =

3 21212

10012

28 87

λ λy z

. .= = =

375028 87

129 9

CHARACTERISTIC STRENGTH AND MODULUS OF ELASTICITY OF CLASS C16 TIMBERValues in Nmm−2

Compressive strength parallel 5% modulus ofto grain, fc,0,k elasticity, E0,05

17 5400

RELATIVE SLENDERNESS RATIO

λ λ

λπ πrel,y rel,z

y c,0,k .

.,

= = = × =fE0 05

129 9 175400

2 32

Since λ rel,y and λ rel,z > 0.3, use the more stringent of equations 11.40 and 11.41.

Example 11.4 Analysis of column resisting an axial load (EC 5)A mechanically graded timber column of strength class C16 consists of a 100 mm square section which is restrainedat both ends in position but not in direction. Assuming that the service conditions comply with Service Class 2 andthe actual height of the column is 3.75m, calculate the design axial long term load that the column can support.

9780415467193_C11 9/3/09, 4:51 PM478

479


Example 11.5 Analysis of an eccentrically loaded column (EC 5)Check the adequacy of the column in Example 11.4 to resist a long-term design axial load of 10 kN applied 35 mmeccentric to its y–y axis.

SLENDERNESS RATIO

λ y = λ z =

lef

i = 129.9 (see Example 11.4)

CHARACTERISTIC STRENGTHS AND MODULUS OF ELASTICITYValues in Nmm−2 for machine graded timber of strength class C16

Bending parallel Compression parallel Modulus of elasticityto grain, fm,k to grain fc,0,k 5-percentile value E0,05

16 17 5400

AXIAL LOAD CAPACITYky = kz = 0.5(1 + βc(λrel,y − 0.3) + λ2

rel,y)

= 0.5(1 + 0.2(2.32 − 0.3) + 2.322) = 3.37

k k

k kc,y c,zy y rel,y

( )

. ( . . )

.= =+ −

=+ −

=1 1

3 37 3 37 2 320 17

2 2 2 2λ

Design compressive strength parallel to grain is given by

f k

kc,0,d mod

c,0,k

M

N mm . .

. = = × = −

γ0 8

17

1 310 46 2

whereγM = 1.3, obtained from Table 11.4kmod = 0.8 (service class 2 and medium-term loading)

By inspection use either equation 11.40 or 11.41. Substituting into equation 11.40 gives

σc,0,d ≤ kc,yfc,0,d

σc,0,d = 0.17 × 10.46 = 1.77 Nmm−2

Hence, axial load capacity of column, N, is given by

N = σc,0,dA = 1.77 × 104 = 17.7 × 103 N = 17.7 kN

9780415467193_C11 9/3/09, 4:51 PM479


480

Design of columns

Example 11.5 continuedBUCKLINGDesign compression stress, σc,0,d, is

σc,0,d

design axial load

AN mm

= =

×= −10 10

101

3

42

Design bending stress about y–y axis, σm,y,d, is

σm,y,d

y

y

N mm

. = =

××

= −M

W

350 10

167 102 10

3

32

Design bending strength about the y–y axis, fm,d, is

f k k

fm,d h mod

m,k

M

N mm . . .

. = = × × = −

γ1 08 0 8

16

1 310 63 2 (Example 11.4)

Design compression strength, fc,0,d, is

f k

kc,0,d mod

c,0,k

M

N mm . .

. = = × = −

γ0 8

17

1 310 46 2

Design bending stress about the z–z axis, σm,z,d = 0. Compression factor is kc,y = 0.17 (see Example 11.4). Check thesuitability of the column by using the more stringent of equations 11.48 and 11.49. By inspection equation 11.48 ismore critical. Substitution gives

σ σ σc,0,d

c,y c,0,d

m,y,d

m,y,dm

m,z,d

m,z,dk f fk

f + + ≤ 1

1

0 17 10 46

2 1

10 63

00 56 0 20 0 76 1

. .

.

. . . .

×+ + = + = <k

fm

m,z,d

OK

Therefore 100 × 100 column is adequate in buckling.

LATERAL TORSIONAL STABILITYlef = l = 3750 mm

σm,crit

ef

.

.

. .,= =

××

× × =0 78 0 78 100

100 37505 4 10 112 3

2

0 05

23b

hE

l

λ

σrel,mm,k

m,crit

.

.= = =f 16

112 30 4

Since λrel,m < 0.75, kcrit = 1Check lateral torsional stability using equation 11.50

σ σm,d

crit m,y,d

c,0,d

c,z c,0,dk f k f

+ ≤

2

1

2 11 0 10 36

10 17 11 3

0 04 0 52 0 56 12

.. .

. .

. . . ×

+×

= + = < OK

Hence the column is also adequate in lateral torsional buckling.

9780415467193_C11 9/3/09, 4:51 PM480

Permissible stress and load factor design

481

The purpose of this appendix is to illustrate thesalient features and highlight essential differencesbetween the following philosophies of structuraldesign:

1. permissible stress approach, i.e. elastic design;2. load factor approach, i.e. plastic design.

The reader is referred to Chapter 2 for revisionof some basic concepts of structural analysis.

Example A.1Consider the case of a simply supported, solid rectangular beam (Fig. A.1), depth (d ) 200 mm, span (l ) 10 m andsubject to a uniformly distributed load (w) of 12 kN m−1. Calculate the minimum width of beam (b) using permissiblestress and load factor approaches to design assuming the following:

σyield = 265 N mm−2

Factor of safety (f.o.s.) = 1.5

Fig. A.1 Simply supported beam.

Appendix A

Permissible stress andload factor design

9780415467193_D01 9/3/09, 4:55 PM481


482

It can therefore be seen that for the case of a simply supported beam, provided all the factors are taken into account,both approaches will give the same result and this will remain true irrespective of the shape of the section.

PERMISSIBLE STRESS APPROACHElastic design moment, Me

M

wle = =

×

2 2

812 10

8(Table 2.4 )

= 150 kN m

= 150 × 106 N mm

Moment of resistance, Mr

Permissible stress, σperm, is

σ

σperm

yield

f.o.s.= =

.2651 5

M r = σperm Z

where Z is the elastic section modulus = bd 2/6(Table 2.4). Hence

M

br 1.5= ×

( )

265 2006

2

= 1.178 × 106b N mm−2

Breadth of beamAt equilibrium, M e = M r

150 × 106 = 1.178 × 106b

Hence breadth of beam, b, is

b = 150/1.178 = 127 mm

LOAD FACTOR APPROACHPlastic design moment, Mp

Plastic section modulus, S =

bd 2

4

Elastic section modulus, Z =

bd 2

6(Table 2.4 )

Shape factor (s.f.) = S/ZHence

s.f. = (bd 2/4)/(bd 2/6) = 1.5

Load factor = s.f. × f.o.s.

= 1.5 × 1.5 = 2.25

Actual load (w) = 12 kN m−1

Factored load (w ′) = 12 × 2.25 = 27 kN m−1

M

w lp = =

×

′ 2 2

827 10

8

= 337.5 kN m = 337.5 × 106 N mm


M S

bdr yield= =

×

σ

2654

2

=

×= ×

.

265 2004

2 65 102

6bb

Breadth of beamAt equilibrium, Mp = M r

337.5 × 106 = 2.65 × 106b


b = 337.5/2.65 = 127 mm

Example A.1 continued

9780415467193_D01 9/3/09, 4:55 PM482


483

PERMISSIBLE STRESS APPROACHElastic design moment, Me

M

wle = =

×

2 2

1212 10

12

= 100 kN m

= 100 × 106 N mm


M Z

br perm= = ×

( ).

σ2651 5

2006

2

= 1.178 × 106b N mm−2

LOAD FACTOR APPROACHPlastic design moment, Mp

Shape factor (s.f.) = S/Z

Hences.f. = (bd 2/4)/(bd 2/6)

= 1.5Load factor = s.f. × f.o.s.

= 1.5 × 1.5 = 2.25Actual load (w) = 12 kN m−1

Factored load (w ′) = 12 × 2.25= 27 kN m−1

M

w lp = =

×

′ 2 2

1627 10

16

= 168.7 kN m = 168.7 × 106 N mm


Mr = σyieldS = 265 ×

bd 2

4

= 265 ×

2004

2b = 2.65 × 106b

Example A.2Repeat Example A.1 but this time assume that the beam is built in at both ends as shown in Fig. A.2.

Fig. A.2 Beam with built-in supports.

9780415467193_D01 9/3/09, 4:55 PM483


484

Example A.2 continuedBreadth of beamAt equilibrium, Me = Mr

100 × 106 = 1.178 × 106b


b = 100/1.178 = 85 mm

Breadth of beamAt equilibrium, Mp = Mr

168.7 × 106 = 2.65 × 106b


b = 168.7/2.65 = 75 mm

Hence, it can be seen that the load factor approach gives a more conservative estimate for the breadth of the beam,a fact which will be generally found to hold for other sections and indeterminate structures.

The basic difference between these two approaches to design is that while the permissible stress method modelsbehaviour of the structure under working loads, and realistic predictions of behaviour in service can be calculated, theload factor method only models failure, and no information on behaviour in service is obtained.

9780415467193_D01 9/3/09, 4:55 PM484

485

D dT

y

t

y

t

B

x xD dxx

y b

Universal Beam

y b

Universal Column

B

T

Appendix B

Dimensions andproperties of steeluniversal beamsand columns

9780415467193_D02 9/3/09, 4:54 PM485

486

Tab

le B

1D

imen

sion

s an

d pr

oper

ties

of

stee

l un

iver

sal

beam

s (s

truc

tura

l se

ctio

ns t

o B

S 4

: P

art

1)

Dim

ensi

ons

Pro

pert

ies

Des

igna

tion

Dep

thW

idth

Thi

ckne

ssR

oot

Dep

thR

atio

s fo

rS

econ

d m

omen

tR

adiu

sE

last

icP

last

icB

uckl

ing

Tor

sion

alW

arpi

ngT

orsi

onal

Are

aof

ofra

dius

betw

een

loca

l bu

cklin

gof

are

aof

gyr

atio

nm

odul

usm

odul

uspa

ram

eter

inde

xco

nsta

ntco

nsta

ntof

Ser

ial

Mas

sse

ctio

nse

ctio

nW

ebF

lang

efil

lets

sect

ion

size

per

Fla

nge

Web

Axi

sA

xis

Axi

sA

xis

Axi

sA

xis

Axi

sA

xis

met

reD

Bt

Tr

db/

Td/

tx–

xy–

yx–

xy–

yx–

xy–

yx–

xy–

yu

xH

JA

(mm

)( k

g)(m

m)

(mm

)(m

m)

(mm

)(m

m)

(mm

)(c

m4 )

(cm

4 )(c

m)

(cm

)(c

m3 )

(cm

3 )(c

m3 )

(cm

3 )(d

m6 )

(cm

4 )(c

m2 )

914×

419

388

920.

542

0.5

21.5

36.6

24.1

799.

15.

7437

.271

900

045

400

38.1

9.58

1560

021

6017

700

334

00.

884

26.7

88.7

173

049

434

391

1.4

418.

519

.432

.024

.179

9.1

6.54

41.2

625

000

3920

037

.89.

4613

700

1870

1550

02

890

0.88

330

.175

.71

190

437

914×

305

289

926.

630

7.8

19.6

32.0

19.1

824.

54.

8142

.150

500

015

600

37.0

6.51

1090

010

1012

600

160

00.

867

31.9

31.2

929

369

253

918.

530

5.5

17.3

27.9

19.1

824.

55.

4747

.743

700

013

300

36.8

6.42

951

087

210

900

137

00.

866

36.2

26.4

627

323

224

910.

330

4.1

15.9

23.9

19.1

824.

56.

3651

.937

600

011

200

36.3

6.27

826

073

89

520

116

00.

861

41.3

22.0

421

285

201

903.

030

3.4

15.2

20.2

19.1

824.

57.

5154

.232

600

09

430

35.6

6.06

721

062

18

360

983

0.85

346

.818

.429

325

683

8×29

222

685

0.9

293.

816

.126

.817

.876

1.7

5.48

47.3

340

000

1140

034

.36.

277

990

773

916

01

210

0.87

35.0

19.3

514

289

194

840.

729

2.4

14.7

21.7

17.8

761.

76.

7451

.827

900

09

070

33.6

6.06

665

062

07

650

974

0.86

241

.615

.230

724

717

683

4.9

291.

614

.018

.817

.876

1.7

7.76

54.4

246

000

779

033

.15.

905

890

534

681

084

20.

856

46.5

13.0

222

224

762×

267

197

769.

626

8.0

15.6

25.4

16.5

685.

85.

2844

.024

000

08

170

30.9

5.71

623

061

07

170

959

0.86

933

.211

.340

525

117

376

2.0

266.

714

.321

.616

.568

5.8

6.17

48.0

205

000

685

030

.55.

575

390

513

620

080

70.

864

38.1

9.38

267

220

147

753.

926

5.3

12.9

17.5

16.5

685.

87.

5853

.216

900

05

470

30.0

5.39

448

041

25

170

649

0.85

745

.17.

4116

118

868

6×25

417

069

2.9

255.

814

.523

.715

.261

5.1

5.40

42.4

170

000

662

028

.05.

534

910

518

562

081

00.

872

31.8

7.41

307

217

152

687.

625

4.5

13.2

21.0

15.2

615.

16.

0646

.615

000

05

780

27.8

5.46

437

045

45

000

710

0.87

135

.56.

4221

919

414

068

3.5

253.

712

.419

.015

.261

5.1

6.68

49.6

136

000

518

027

.65.

383

990

408

456

063

80.

868

38.7

5.72

169

179

125

677.

925

3.0

11.7

16.2

15.2

615.

17.

8152

.611

800

04

380

27.2

5.24

348

034

64

000

542

0.86

243

.94.

7911

616

061

0×30

523

863

3.0

311.

518

.631

.416

.553

7.2

4.96

28.9

208

000

1580

026

.17.

226

560

1020

746

01

570

0.88

621

.114

.378

830

417

961

7.5

307.

014

.123

.616

.553

7.2

6.50

38.1

1520

011

400

25.8

7.08

491

074

35

520

114

00.

886

27.5

10.1

341

228

149

609.

630

4.8

11.9

19.7

16.5

537.

27.

7445

.112

500

09

300

25.6

6.99

409

061

04

570

937

0.88

632

.58.

0920

019

061

0×22

914

061

7.0

230.

113

.122

.112

.754

7.3

5.21

41.8

112

000

451

025

.05.

033

630

392

415

061

20.

875

30.5

3.99

217

178

125

611.

922

9.0

11.9

19.6

12.7

547.

35.

8446

.098

600

393

024

.94.

963

220

344

368

053

60.

873

34.0

3.45

155

160

113

607.

322

8.2

11.2

17.3

12.7

547.

36.

6048

.987

400

344

024

.64.

882

880

301

329

047

00.

8737

.92.

9911

214

410

160

2.2

227.

610

.614

.812

.754

7.3

7.69

51.6

7570

02

910

24.2

4.75

251

025

62

880

400

0.86

343

.02.

5177

.212

953

3×21

012

254

4.6

211.

912

.821

.312

.747

6.5

4.97

37.2

7620

03

390

22.1

4.67

280

032

03

200

501

0.87

627

.62.

3218

015

610

953

9.5

210.

711

.618

.812

.747

6.5

5.60

41.1

6670

02

940

21.9

4.60

247

027

92

820

435

0.87

530

.91.

9912

613

910

153

6.7

210.

110

.917

.412

.747

6.5

6.04

43.7

6170

02

690

21.8

4.56

230

025

72

620

400

0.87

433

.11.

8210

212

992

533.

120

9.3

10.2

15.6

12.7

476.

56.

7146

.755

400

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9780415467193_D02 9/3/09, 4:54 PM486

487

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9780415467193_D02 9/3/09, 4:54 PM487

488

Tab

le B

2D

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sion

s an

d pr

oper

ties

of

stee

l un

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sal

colu

mns

(st

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sec

tion

s to

BS

4:

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t 1)

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260.

425

5.9

10.5

17.3

12.7

200.

37.

4019

.114

300

485

011

.26.

521

100

379

123

057

50.

849

14.4

0.71

610

411

473

254.

025

4.0

8.6

14.2

12.7

200.

38.

9423

.311

400

387

011

.16.

4689

430

598

946

20.

849

17.3

0.55

757

.392

.920

3×20

386

222.

320

8.8

13.0

20.5

10.2

160.

95.

0912

.49

460

312

09.

275.

3285

129

997

945

60.

8510

.20.

317

138

110

7121

5.9

206.

210

.317

.310

.216

0.9

5.96

15.6

765

02

540

9.16

5.28

708

246

802

374

0.85

211

.90.

2581

.591

.160

209.

620

5.2

9.3

14.2

10.2

160.

97.

2317

.36

090

204

08.

965.

1958

119

965

230

30.

847

14.1

0.19

546

.675

.852

206.

220

3.9

8.0

12.5

10.2

160.

98.

1620

.15

260

177

08.

905.

1651

017

456

826

40.

848

15.8

0.16

632

.066

.446

203.

220

3.2

7.3

11.0

10.2

160.

99.

2422

.04

560

154

08.

815.

1144

915

149

723

00.

846

17.7

0.14

222

.258

.815

2×15

237

161.

815

4.4

8.1

11.5

7.6

123.

56.

7115

.22

220

709

6.84

3.87

274

91.8

310

140

0.84

813

.30.

0419

.547

.430

157.

515

2.9

6.6

9.4

7.6

123.

58.

1318

.71

740

558

6.75

3.82

221

73.1

247

111

0.84

816

.00.

0306

10.5

38.2

2315

2.4

152.

46.

16.

87.

612

3.5

11.2

20.2

126

040

36.

513.

6816

652

.918

480

.90.

837

20.4

0.02

144.

8729

.8

9780415467193_D02 9/3/09, 4:54 PM488

489

Appendix C

Buckling resistanceof unstiffened webs

Substitution gives a slenderness of

λp

yw=+

.( )

0 659 1

2

b nk dpEt

The buckling resistance, PX, can be written as

Pb nk dp

Et

PX

yw

bw=+

.

.( )

0 65

0 659 12

Re-arranging gives

P

tp

b nk dPX

ywbw=

+

.

( )

27 2 275

1

Comparison of the above equation with availabletest data highlights the approximations made inthe above formulation and led to a reduction ofthe factor 27.2 by 8%, to 25.0.

Letting

ε =

275pyw

results in

P

t

b nk dPX bw

( )=

+25

1

ε

which represents the equation given in BS 5950–1:2000 for an unstiffened web.

When the applied load or reaction is less than0.7d from the end of the member, the bucklingresistance of an unstiffened web is reduced by thefactor

a dd

e + ..0 7

1 4

where ae < 0.7d and is the distance from the loador reaction to the end of the member.

The previous version of BS 5950–1 calculated thebuckling resistance of unstiffened webs assumingthe web behaved as a strut with a slenderness of2.5d /t. Comparison with test results suggested thatthis approach could, in a limited number of cases,lead to unconservative estimates of the web’s buck-ling resistance. It was therefore decided in BS 5950–1:2000 to revise the design approach and base theweb’s buckling resistance on the well-known theoryof plate buckling, which represents the behaviourmore realistically compared to the previous assump-tion of the web acting as a strut.

The buckling resistance, PX, of a plate is givenby

PX = ρPbw

where ρ is a reduction factor based on the effectivewidth concept of representing the inelastic postbuckling of plates, and Pbw is the bearing capacity.The reduction factor can, approximately, be givenby1

ρ =

0 65.λp

where the slenderness of the plate, λp, is given by

λ p

bw

elastic

=P

PRepresenting an unstiffened web as a plate,BS 5950–1:2000 defines the bearing capacity, Pbw,as

Pbw = (b1 + nk)tpyw

Pelastic is the elastic buckling load of the web. As-suming that the web is restrained by the flanges ofthe section and the web behaves as a long plate,the elastic buckling load is given by2

P

Etv delastic

( )=

−2

3 1

3

2

π

9780415467193_D03 9/3/09, 4:54 PM489

490

Buckling resistance of unstiffened webs

References1. Bradford, M.A. et al., Australian Limit State Design

Rules for the Stability of Steel Structures, First NationalStructural Engineering Conference, pp 209–216, Mel-bourne 1987.

2. Timoshenko, S.P. and Gere, J.M., Theory of ElasticStability, McGraw-Hill, 1961.

9780415467193_D03 11/3/09, 11:19 AM490

491

Appendix D

Second momentof area of acomposite beam

Deflections of composite beams are normallycalculated using the gross value of the secondmoment of area of the uncracked section, Ig. Thisappendix derives the formula for Ig given in section4.10.3.6.

Consider the beam section shown in Fig. D1which consists of a concrete slab of effective width,Be, and depth, Ds, acting compositely with a steelbeam of cross-sectional area, A, and overall depth,D.

Assuming the modular ratio is αe, the transformedarea of concrete slab is (Be/αe)Ds. Taking momentsabout x–x, the distance between the centroids ofthe concrete slab and the steel beam A, is

A

( )

( )=

+

+

=+

+

AD D

AB D

A D DA B D

2 2

s

e s

e

e s

e e s

α

αα2

(D1)

The second moment of area of the composite sec-tion, Ig, is then

I I

B DA

D D B Dg s

s3

e

s e s

e

2

12 2 2= + + +

−

+ e

α αA A

2

(D2)

where Is is the second moment of area of the steelsection.

Making (Ds + D)/2 the subject of equation (D1)and substituting into (D2) gives

I I

B DA

A B DAg s

s3

e

e s e

12= + +

+

−

/ e

αα

A A

2

+

B De s

e

2

αA (D3)

Simplifying and substituting (D1) into (D3) gives

I I

B D AB DA

A D DA B Dg s

s3

e

e2

s2

e2

e s

e e s12= + +

−+

( )

( )e

α ααα2

2

2

+

++

( )

( )B D A D D

A B De s

e

e s

e e sααα2

2

(D4)

Collecting terms and simplifying obtains theequation given for the gross value of the secondmoment of area of the uncracked composite sectionquoted in section 4.10.3.6

I I


s3

e

e s s

e e s12= + +

++

( )

( )e

α α

2

4(D5)

Fig. D1

Be/αe

Ds

DTransformedarea of slab

Steel beam of cross-sectional area A

D/2 + Ds/2

x

Be

x

−y

9780415467193_D04 9/3/09, 4:53 PM491

9780415467193_D04 9/3/09, 4:53 PM492

493

References and further reading

BS 8007: 1987: Code of practice for the designof concrete structures for retainingaqueous liquids

BS 8110: Structural use of concrete; Part 1:Code of practice for design andconstruction, 1997; Part 2: Code ofpractice for special circumstances,1985; Part 3: Design charts for singlyreinforced beams, doubly reinforcedbeams and rectangular columns,1985

BS 8500–1: 2006: Concrete – complementaryBritish Standard to BS EN 206–1;Part 1: Method of specifying andguidance for the specifier

CP3: 1972: Code of basic design data forthe design of buildings; Chapter V:Part 2: Wind loads

CP114: 1969: Structural use of reinforcedconcrete in buildings

EUROCODES, NATIONAL ANNEXES, PD ANDEUROPEAN STANDARDSBS EN 1990, Basis of structural design, 2002Eurocode:BS EN 1991, Actions on Structures. GeneralEurocode 1: actions. Part 1–1: Densities, self

weight and imposed loads, 2002;Part 1–4 Wind loads, 2005

BS EN 1992, Design of concrete structures –Eurocode 2: Part 1–1: General rules and rules

for buildings, 2004BS EN1993, Design of steel structures –Eurocode 3: Part 1–1: General rules and rules

for buildings, 2005; Part 1–5:Plated structural elements, 2006;Part 1–8: Design of joints, 2005

BS EN1995, Design of Timber Structures –Eurocode 5: Part 1–1: Common Rules and

rules for buildings, 2004

ReferencesBRITISH STANDARDSBS 4–1: 2005: Structural steel sections;

Part 1: Specification for hot-rolledsections

BS 449: 1996: Specification for the use ofstructural steel in buildings: Part 2:Metric units

BS 648: 1964: Schedule of weights of build-ing materials

BS 4729: 2005: Clay and calcium silicatebricks of special shapes and sizes –recommendations

BS 5268–2: 2002: Structural use of timber –Part 2: Code of practice for per-missible stress design, materials andworkmanship

BS 5400: Steel, Concrete and CompositeBridges; Part 3: Code of Practice fordesign of steel bridges, 2000; Part 4:Code of Practice for design of con-crete bridges, 1990

BS 5628: 2005: Code of practice for use ofmasonry; Part 1: Structural use ofunreinforced masonry; Part 2:Structural use of reinforced andprestressed masonry; Part 3: Mater-ials and components, design andworkmanship

BS 5950–1: 2000: Structural use of steelworkin buildings; Part 1: Code of prac-tice for design – rolled and weldedsections

BS 6399: Design loading for buildings; Part 1:Code of practice for dead andimposed loads, 1996; Part 2: Codeof practice for wind loads, 1997; Part3: Code of practice for imposed roofloads, 1988

Appendix E

References andfurther reading

9780415467193_D05 11/3/09, 11:20 AM493


494

BS EN 1996, Design of masonry structuresEurocode 6: – Part 1–1: General rules for

reinforced and unreinforcedmasonry structures, 2005

NA to BS UK National Annex to Euro-EN 1990: code 0: Basis of structural

design, 2002NA to BS UK National Annex to Euro-EN 1991–1–1: code 1: Actions on structures.

General actions. Densities,self-weight and imposed loads,2002

NA to BS UK National Annex to Euro-EN 1992–1–1: code 2: Design of concrete

structures – Part 1–1: Generalrules and rules for buildings,2004

NA to BS UK National Annex to Euro-EN 1993: code 3: Design of steel structures

– Part 1–1: General rules andrules for buildings, 2008; Part1–5: Plated structural elements,2006

NA to BS UK National Annex toEN 1995–1–1: Eurocode 5: Design of Timber

Structures – Part 1–1: General– Common Rules and Rules forBuildings, 2004

NA to BS UK National Annex to Euro-EN 1996–1–1: code 6: Design of masonry

structures – Part 1–1: Generalrules for reinforced andunreinforced masonry structures,2005

PD 6678: Background paper to the UKNational Annex to BS EN1992–1, 2006

BS EN 206–1: 2000, Concrete – Part 1:Specification, performance,production and conformity

BS EN 336: 2003, Structural timber – sizes,permitted deviations

BS EN 338: 2003, Structural timber –strength classes

BS EN 771: 2003, Specification for masonryunits; Part 1: Clay masonryunits; Part 3: Aggregate concretemasonry units (dense and light-weight aggregates)

BS EN 772–1: 2000, Methods of tests formasonry units. Determination ofcompressive strength

BS EN 845–1: 2003, Specification for ancillarycomponents for masonry –

Part 1: Ties, tension straps,hangers and brackets

BS EN 10002: 2001, Metallic materials –Tensile testing – Part 1:Methods of test at ambienttemperature

BS EN 12390–3: 2002, Testing hardened concrete– Part 3: Compressive strengthof test specimens

BS EN 14081: 2005, Timber structures –Strength graded structural timberwith rectangular cross section –Part 1: General requirements

Further readingBond, A.J. et al., How to design concrete structures

using Eurocode 2, The Concrete Centre, 2006Curtin, W.G. et al., Structural masonry designers’

manual, 3rd edition, Oxford, Blackwell, 2006Gardner, L. and Nethercot, D.A., Designers’ guide

to EN 1993–1–1, Eurocode 3: Design of steel struc-tures, general rules and rules for buildings, London,Thomas Telford, 2005

Higgins, J.B. and Rogers, B.R., Designed anddetailed (BS 8110: 1997), Crowthorne, BritishCement Association, 1998

Institution of Structural Engineers and TheConcrete Society, Standard method of detailingstructural concrete, London, Institution of Struc-tural Engineers/Concrete Society, 1989

Institution of Structural Engineers and The Con-crete Society, Standard method of detailing structuralconcrete – a manual for best practice, London,Institution of Structural Engineers/ConcreteSociety, 2006

Institution of Structural Engineers, Manual forthe design of timber building structures to Eurocode5, London, Institution of Structural Engineers,2007

Institution of Structural Engineers, Manual for thedesign of plain masonry in building structures toEurocode 6, London, Institution of StructuralEngineers, 2008

Mosley, B. et al., Reinforced concrete design toEurocode 2, 6th edition, Basingstoke, Hampshire,Palgrave Macmillan, 2007

Narayanan, R.S. and Beeby, A.W., Designers’ guideto EN 1992–1–1 and EN 1992–1–2, Eurocode 2:Design of Concrete Structures. General rules and rulesfor buildings and fire design, London, ThomasTelford, 2005

Narayanan, R.S. and Goodchild, C.H., ConciseEurocode 2 for the design of in-situ concrete framed

9780415467193_D05 11/3/09, 11:21 AM494

495


buildings to BS EN 1992–1–1: 2004 and its UKNational Annex: 2005, Blackwater, Concrete Cen-tre, 2006

Ozelton, E.C. and Baird, J.A., Timber designers’manual, 3rd edition, Oxford, Blackwell Science,2002

Reynolds et al., Reynolds’s reinforced concretedesigner’s handbook, 11th edition, London, Taylor& Francis, 2008

Threlfall, A.J., Designed and detailed (Eurocode 2:2004), Concrete Society/British Cement Associ-ation, Blackwater, 2009

9780415467193_D05 9/3/09, 4:53 PM495


496

9780415467193_D05 9/3/09, 4:53 PM496

497

Index

Index

Asv/sv ratios 51, 54, 332Actions 310, 317, 378, 435, 460

characteristic 317, 378, 436, 461combination expressions 319, 378design 318, 378, 436, 461frequent 319partial safety factor 318, 379, 437, 461permanent 317, 378, 436, 460quasi-permanent 319variable 319, 378, 436, 460ψ0, ψ2 319, 465

Application rules 311, 376Axes 377, 464

Basis of design 4, 33, 377, 436, 460Beam design 24, 94Beam theory 24–26Beams (see also flexural members) 44, 151, 287,

327, 380anchorage length 60, 339bending 45, 152, 264, 287, 327, 381, 455, 464bending and shear 156, 160, 381bond 340buckling factor 398, 399buckling resistance 399cantilever 153, 159, 161, 176, 393continuous 70, 320, 343curtailment 60, 90, 342deflections 22, 57, 85, 98, 153, 159–162, 287,

299, 337, 384, 464design charts 49, 70doubly reinforced 44, 67, 328effective length/span 57, 98, 167, 176, 181, 287,

338, 362equivalent slenderness 174flange buckling 152, 382high shear 156, 392L-sections 44, 71lap length 60, 343lateral buckling 290, 466lateral torsional buckling 152, 167–177, 184, 398,

406laterally restrained 156, 380lever arm 47local buckling 152

low shear 156, 158, 381, 385moment capacity 156, 159, 381–382over-reinforced 46preliminary sizing 57reinforcement areas 49, 59, 339reinforcement details 53, 339section classification 154, 380shear 50, 155, 291, 330, 381, 467shear area 155, 381shear buckling 152, 382shear capacity 155, 381singly reinforced 45, 327spacing of reinforcement 52, 59, 334, 339span/effective depth ratio 57, 58, 337stiffener design 166, 397T-sections 44, 71, 80under-reinforced 46, 327universal beams, dimensions and properties

485–487web bearing 152, 162web buckling 152, 162, 489web crippling 383web crushing 382web failure 153, 383

Bending moments (see also structural analysis)coefficients 71, 74, 76, 78, 106, 109, 269equilibrium equations 18formulae 21

Bending strength 169–171, 173Blocks 242, 244, 439

aggregate 242aircrete 242cellular 243, 438compressive strengths 243hollow 243, 438shape factor 438solid 243work sizes 243

Bolted connections (see also HSFG) 218, 418bearing capacity 220, 420block shear/tearing 223, 425bolt strength 221, 222, 418clearance 219, 419design 220, 421double shear 220

9780415467193_D06_index 9/3/09, 4:52 PM497

Index

498

shear and tension 222shear capacity 220, 221, 419tension capacity 221

Bracing 241Bricks 241

classification 243clay 242commons 242coordinating size 241durability 242, 441engineering 243facing 242frogged 241manufacture 241solid 241soluble salt content 242specification 242work size 241

Brickwork and blockwork 247, 260characteristic compressive strength 247–248, 440characteristic flexural strength 263, 455

British Standards 4, 10, 493

Characteristic actions 317Characteristic loads 6, 9–12, 247Characteristic strengths 6, 34, 151, 247, 265, 440,

455, 462Columns (see also compression members) 26, 128,

361axial load and bending 133, 185, 300, 363, 405,

477axially loaded 26, 132, 177, 299, 403, 477baseplates 407, 417biaxial bending 137, 184, 364braced and unbraced 129, 183, 363buckling length 404buckling resistance 184, 403–404classification 129, 183, 185, 362, 367, 369, 380,

409, 412compression resistance 403definition 128design 26, 128, 191, 298design charts 133, 363eccentrically loaded 132eccentricities 316, 363–364effective height/length 130, 182, 250, 299, 362,

405, 443, 477end restraints 13, 130, 182, 250, 265–267, 269,

299, 362equivalent uniform moment factor 406failure mechanisms 7, 129, 177, 184imperfection factor 404links 138, 365longitudinal reinforcement 138, 365preliminary sizing 133, 366reduction factor 405reinforcement details 138, 365short-braced 128, 131, 141slender 129

slenderness ratio 26, 129, 177, 181, 250, 298,361–363, 404

uniaxial bending 133, 137Composite beams 201

deflection 211effective breadth 203longitudinal shear capacity 211moment capacity 204, 209, 210shear capacity 207shear connectors 207

Composite construction 199advantages 200beams 201columns 191floor slabs 199, 201

Compression members (see also columns) 177, 199,201, 298, 403

axially loaded 177, 299, 403axial load and bending 184, 300, 405baseplates 407, 417bracing 150buckling resistance check 184, 189, 192cased columns 191compressive strength 178–181cross-section capacity check 184, 192, 405effective length/height 182, 300, 477end conditions 182, 300equivalent slenderness 189load eccentricities 188load sharing 284, 303, 463, 470non-dimensional slenderness 404radius of gyration 26, 177, 299, 404, 477,

486–488relative slenderness ratio 477strut selection table 178strut 177simple construction 150, 188, 407slenderness ratio 26, 177, 298universal columns, dimensions and properties 485,

488Compressive strength 34, 178–181, 242–243,

247–248, 281, 283, 317, 437, 477Conceptual design 4Concrete design 31, 314Connections 218, 418

beam-to-beam 219, 228beam-to-column 219, 224, 232, 235, 424, 426,

428, 429beam splice 228bearing 220, 420bolted 219, 419bracket-to-column 227combined shear and tension 222, 223double shear 220, 422ductility 424effective length 235effective throat size 234, 421failure modes 220, 223HSFG/preloaded bolts 223, 420

9780415467193_D06_index 9/3/09, 4:52 PM498

499

Index

ordinary (black) bolts 220, 418single shear 220, 419splice 421–422tension 221web cleat beam to column 224, 428, 429welded 234, 420welded end plate to beam 232, 423–424, 426

Construction control (see also execution control)normal 249special 249

Continuous design 150Cover 37–41, 43, 44, 324–327Cracking 41, 59, 99, 103, 326, 339, 355

Damp proof courses 240Deflection 22, 57, 98, 159, 161, 289, 337–339, 384,

464–465deformation factor 465final 465instantaneous 465modification factor 57, 58net, final 465span/effective depth ratios 57, 58, 337

Density 11, 12, 283Design loads (see also Actions) 35, 150, 246Design philosophy 5

Limit state 5–8Load factor 5, 481Permissible stress 5, 145, 279, 481

Design process 4, 10, 154, 334Design strengths 8, 34, 35, 151, 234, 247, 317, 378,

437, 455Detailed design 4Durability 37, 149, 244, 325, 441, 460Duration of loading 283, 463

Equivalent uniform moment factor 172, 185Euler 177Eurocodes 307

benefits 109design philosophy 314, 375, 434, 458EN 309EN 1990 318ENV 309Eurocode 1 317Eurocode 2 314Eurocode 3 375Eurocode 5 458Eurocode 6 434implementation 312maintenance 312National Annex 311NCCI 312Implementation 312Scope 309

Execution control (see also construction control) 440class 1 440class 2 440

Exposure classes 38

Fire resistance 44, 150, 324Flat slabs 94

shear reinforcement 95–97Flexural members 287, 464

bearing 287, 468bending 287, 464bending deflection 289–290deflection 287, 464design 287, 464effective length/span 287, 467lateral buckling 290, 466notched ends 284, 468relative slenderness ratio 466shear 291, 467shear deflection 289, 290vibration 465wane 291

Flexural strength 265Floors (see also slabs)

concrete frames 93in-situ concrete 199metal deck 201precast concrete 199steel frames 199

Floor joists 293, 469Foundations 115

design 116face shear 117, 120failure 115pad 115, 116piled 116punching shear 117, 119raft 116strip 115transverse shear 117, 120

Geometrical properties of sawn timber 288Grade stresses 281, 283

Hardwood 279Hooke’s law 24HSFG bolts 223, 420

bearing capacity 223proof load 223shear capacity 223slip factor 223, 420slip resistance 223, 420ultimate tensile strength 418yield strength 418

Internal forces 310Internal moments 310

k factors (see also modification factors)bending 464compression 468deformation 465depth 463instability 466

9780415467193_D06_index 9/3/09, 4:52 PM499

Index

500

lateral buckling 466load sharing 463, 470notched ends 467shear 467

Limit state design 5, 33, 145, 246, 314, 377, 434,458

durability 37, 149, 244, 325, 377, 441, 460fire 44, 150, 324, 434, 458serviceability 6, 33, 149, 317, 337, 378, 460ultimate 6, 33, 149, 150, 317, 378, 460

Load duration classes 284, 463Load paths 9Loading (see also Actions) 9–17, 35, 150, 246

arrangements 36, 320characteristic 6, 9–12, 150combinations 11, 35, 247, 318dead 9design 8, 13, 35, 246destabilising 153, 168high shear 156, 381imposed 10normal 153, 168partial safety factor 6, 11, 34, 35, 151, 247, 249,

317, 318wind 10, 150, 263, 318

Masonryadvantages 240applications 239compressive strength 243, 247–248, 440construction/execution control 249, 440durability 242flexural strength 263, 455soluble salt content 242unit quality 249, 440

Masonry design (see also panel walls) 245, 441capacity reduction factor 249, 441, 442cavity 251, 261, 274, 450, 453design strength 247, 437design procedure 253, 270eccentricity of loading 251, 441effective height 250, 443effective thickness 250, 443enhanced resistance 250, 443laterally loaded walls 263, 455load resistance 251, 441mortar 245, 438narrow wall factor 247, 443shape factor 438simple resistance 250, 266, 443slenderness ratio 243, 402small plan area factor 247, 443vertically loaded walls 246, 441

Mechanical grading 280Material strengths

bolts 221, 222, 223, 418characteristic 6

concrete 34, 317concrete blocks 243design 8fillet welds 234, 421masonry 248, 440mortar 245, 438partial safety factor 6, 34, 151, 247, 249, 317, 318,

379, 419, 441, 463reinforcing steel 34, 317structural steel 145, 151, 379timber 281, 283, 462

Modification factors (see also k factors) 282compression members 285depth 284duration of loading 283load sharing 284moisture content 282notched end 284

Modulus of elasticity 24, 36, 37, 146, 161, 281, 283,317, 323, 380, 442, 462

Mortar 245, 438choice 245composition 244compressive strength 244, 438lime 245properties 244–245proportioning 245selection 244

National Annex 311, 320, 494Notched joists 284, 296, 467, 475

Pad foundations 116, 357Panel walls (see also masonry walls) 263, 455

bending moment coefficients 269cavity wall 274design procedure 270failure criteria 264flexural strength 265free edge 265limiting dimensions 267moments of resistance 268, 455one-way spanning 271, 456orthogonal ratio 268, 455restrained supports 266simple supports 266two-way spanning 264, 272, 457

Partial safety factor 6Loads 8, 11, 35, 151, 247, 318, 379, 437, 461materials 6, 34, 151, 249, 317, 441, 463resistances 380, 419

Perry-Robertson 177, 285PD 6687 362, 494Plastic analysis 25, 45, 150, 152, 154Plastic cross-sections 154, 380Principles 311Properties of concrete and steel 31Properties of iron and steel 145–146

9780415467193_D06_index 9/3/09, 4:52 PM500

501

Index

Punching shear 117, 358Critical perimeter 145–146

Radius of gyration 26, 117, 486–488Rankin’s formula 122Reinforcement areas 49, 54, 59, 99, 101, 124, 138,

339, 365Retaining walls 121

analysis and design 121, 125cantilever 121conterfort 121failure modes 123gravity 121

Ronan Point 4

Section propertiessecond moment of area 25, 212, 288, 290,

486–488section modulus 25, 271–273, 275, 287, 288–289,

456elastic modulus 25, 156, 377, 459, 470, 486–488plastic modulus 26, 156, 381, 486–488

Semi-continuous design 150Section classification 155, 380Service classes 282, 463Shear 50, 98, 117, 152, 155, 207, 220, 290, 330,

350aggregate interlock 50beams 52, 330design concrete shear stress 51design force 52, 331design links 52design resistance 332design stress 50diagonal compression 50diagonal tension 50diameter of links 54dowel action 50face 117, 358failure mechanisms 50inclined bars 51maximum shear stress 50nominal links 52shear span 333slabs 99, 350minimum area of links 334, 350modulus 290, 380, 462punching 117, 358transverse 117, 357resistance of links 51, 332spacing of links 52, 334, 350strut angle 332strut capacity 332truss analogy 330, 331

Shear forcecoefficients 71, 106, 110equilibrium equations 18formulae 22

Shift rule 333, 342Simple design/construction 150, 188Slabs (see also floors) 94, 350

ACI shear stirrups 95analysis and design 97, 100, 104, 105, 109anchorage 351bending 350crack width 99, 352curtailment rules 99, 351flat 95continuous, one-way 105reinforcement areas 99, 101, 350reinforcement details 99, 350ribbed 97shear 98, 350shear hoops 96shear ladders 96solid, one-way 94, 97, 350solid, two-way 109spacing of reinforcement 99, 350stud rails 96

Slenderness ratio 26, 177, 250, 298, 361,404

Small plan area 247, 443Softwoods 279, 281–282Span/effective depth ratios 57–59, 98, 337Steel design 145, 375Steel grades 146, 379Strength classes 280, 282–283, 461–462

General Structural 280–283Special Structural 280–283

Stress blocks 46, 65, 68, 135–136, 327, 328Stress-strain curves 24, 36, 37, 146, 323, 324Stresses (see also timber strengths)

basic 280grade 280–281, 283, 461permissible 282

Structural analysis 17, 71, 76bending moments 22, 290bending moment coefficient for beams 71, 74,

78bending moment and shear force coefficients for

slabs 109bending moments coefficients for walls 269carry over factor 72continuous beams 72, 76deflections 22, 290distribution factors 64elastic 24equilibrium equations 18fixed end moments 71, 72formulae 21moment distribution 71plastic 25shear forces 18, 22, 290shear force coefficients for beams 71stiffness factor 72superposition 23

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Index

502

Stud walling 303construction 303design 303

Support conditions 13, 60, 130, 168, 182, 250, 251,290, 299

Symbols 32, 148, 245, 285, 315, 377, 435, 459

Tension reinforcement 47, 67, 328Timber

applications 279density 283design 279, 458species 280, 282, 462hardwood 279softwood 279, 282strength classes 280, 282–283, 462stiffness 281, 283, 462strengths 462

Ultimate moment of resistance 46, 327

Vibration 465, 472damping coefficient 466fundamental frequency 465

Visual grading 280

Wallscavity 240, 246, 251, 261, 274, 450, 453load-bearing 240narrow brick 247, 443non load-bearing 240panel 263, 455piered 246, 251, 256, 443, 447single-leaf 246, 254, 258, 444small plan area 247, 443

Wall ties 240, 241Wane 290Welded connections 234, 420

butt welds 234correlation factor 421design resistance 420design strength 234effective length 235effective throat size 234, 420fillet weld 234, 421leg length 234shear strength 420

Yield strength (see also design strength) 379Yield stress 24Young’s modulus (see modulus of elasticity)

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